Class 10 Life Science

• Chapter 1 Control And Coordination In Living Organisms MCQs
• Chapter 2 Continuity Of Life MCQs
• Chapter 3 Heredity and Common Genetic Diseases MCQs
• Chapter 4 Evolution And Adaptation MCQs
• Chapter 5 Environment Its Resources And Their Conservation MCQs

WBBSE Chapter 5 Nitrogen Cycle Multiple Choice Questions

Question 1. The bacteria that bring about ammonification is

1. Bacillus ramosus
2. B.thuringiensis
3. B. subtilis
4. None of the above

Answer: 1. Bacillus ramosus

Question 2. The bacteria helping in nitrification is

1. Clostridium sp
2. Nitrosomonas sp
3. Nitrobactersp
4. Pseudomonas

Answer: 2. Nitrosomonas sp

Read and Learn More WBBSE Class 10 Life Science Multiple Choice Questions

Question 3. The absorption of nitrogen in plants occurs in the form of

1. N03
2. N₂
3. H2
4. NH4⁺

Answer: 1. N03

Question 4. The bacteria that cause denitrification is

1. Glucanobacter sp
2. Escherichia sp
3. Nitrobacter sp
4. Pseudomonas sp

Answer: 3. Nitrobacter sp

Question 5. Which of the plants does not have endophytic cyanobacteria?

1. Cycas sp
2. Azolla sp
3. Anthoceros sp
4. Gnetum sp

Answer: 3. Anthoceros sp

Question 6. Which one of the greenhouse gases causes the acidification of soil?

1. Ch4
2. CFC
3. N₂O
4. 03

Answer: 3. N₂O

WBBSE Class 10 Life Science Chapter 5 MCQs

Question 7. Which process increases nitrogen concentration in the atmosphere?

1. Ammonification
2. Nitrification
3. Nitrogen fixation
4. Denitrification

Answer: 4. Denitrification

Question 8. Which of the bacteria is a free-living nitrogen fixer?

1. Escherichia
2. Clostridium
3. Rhizobium
4. Frankia

Answer: 2. Clostridium

Question 9. Which bacteria remains associated with the root valgus?

1. Rhizobium
2. Azospirillum
3. Frankia
4. Bradyrhizobium

Answer: 2. Azospirillum

Question 10. The key enzyme for nitrogen fixation is

1. Dinitrogenase
2. Nitrogenase
3. Nitrate oxidase
4. Nitrate reductase

Answer: 2. Nitrogenase

WBBSE Class 10 Life Science Chapter 5 MCQs

WBBSE Class 10 Life Science Chapter 5 MCQs With Answers

Question 11. The elements required for nitrogenase are

1. Co and mo
2. Fe and mo
3. Fe and co
4. B and mo.

Answer: 2. Fe and mo

Question 12. The chemical process involving the production of NH3 is

1. Haber-bosch process
2. Riemer’s process
3. Helmont’s process
4. Friedel-craft’s reaction

Answer: 1. Haber-bosch process

Question 13. The following cyanobacteria fix nitrogen

1. Stigonema
2. Gleocapsa
3. Anabaena
4. None of these

Answer: 3. Anabaena

Question 14. The following fern is used as a biofertilizer in paddy field

1. Salvinea
2. Azolla
3. Marsilea
4. None of the above

Answer: 2. Azolla

Question 15. The number of ATP molecules required for fixing one molecule of nitrogen is

1. 6 Atp
2. 10 Atp
3. 14 Atp
4. 16 Atp

Answer: 4. 16 Atp

Question 16. The nodule appears red due to the presence of

1. Nodulin
2. Haemoglobin
3. Ferredoxin
4. Leg-haemoglobin

Answer: 4. Leg-haemoglobin

WBBSE Class 10 Life Science Chapter 5 MCQs

Question 18. Rhizobium to legume root is

1. Polysaccharide
2. Nodulin
3. Lectin
4. Protein

Answer: 3. Lectin

Question 19. Leghaemoglobin creates

1. Anaerobic condition for optimum activity of nitrogenase
2. Aerobic condition for optimum activity of nitrogenase
3. Required oxygen concentration for optimum activity of nitrogenase
4. A suitable environment for nodule formation

Answer: 3. Required oxygen concentration for optimum activity of nitrogenase

Question 20. Arrange the following processes of the N₂ cycle in proper sequence: Denitrification, Nitrogen fixation, Ammonification, and Nitrification

1. (2) — (3)— (4) — (1)
2. (1) — (2) — (3) — (4)
3. (1) — (3) — (2) — (4)
4. (4) —(3) —(2) —(1)

Answer: 1. (2) — (3)— (4) — (1)

Question 21. One of the following is a way by which atmospheric nitrogen is converted to nitrates-

1. Reduction
2. Decay
3. Denitrification
4. Lightning

Answer: 4. Lightning

Question 22. Rhizobium bacteria and root nodules of a pea plant an examples of

1. Predation
2. Symbiosis
3. Parasitism
4. Commensalism

Answer: 2. Symbiosis

Question 23. Which of the following phases of the nitrogen cycle Pseudomonas is associated with—

1. Nitrogen fixation
2. Nitrification
3. Denitrification
4. Ammonification

Answer: 3. Denitrification

Life Science Class 10 Chapter 5 Mcqs With Answers

Question 24. Which one of the folio wing microbes takes part in nitrification—

1. Nitrosomonas
2. Azotobacter
3. Pseudomonas
4. Thiobacillus

Answer: 1. Nitrosomonas

WBBSE Chapter 5 Environmental Pollution MCQs

Question 25. In almost all metropolitan cities, the major atmospheric pollutant(s) is/are

1. Oxides of sulfur
2. CO₂ & Co
3. Suspended particulate matter (SPM)
4. Oxides of nitrogen

Answer: 3. Suspended particulate matter (SPM)

Question 26. SPM include

1. Aerosols
2. Raindrops
3. Smog
4. Minute dust particles

Answer: 4. Minute dust particles

Question 27. Which of the following is not a particulate matter?

1. Hydrocarbon dust
2. Lead dust
3. Ozone
4. Smoke

Answer: 3. Ozone

Question 28. The diameter of the rpm is

1. SPM₁₀
2. SPM_g
3. SPM_s
4. SPM₂

Answer: 4. SPM₂

Question 29. Which one of the following is a secondary pollutant?

1. O3
2. Pan
3. Both
4. CO₂

Answer: 3. Both

Question 30. Which one of the following is not a greenhouse gas?

1. CO₂
2. Ch4
3. 03
4. Co

Answer: 4. Co

Life Science Class 10 Chapter 5 Mcqs With Answers

Question 31. The percentage of air pollution that comes from combustion is

1. 20-30%
2. 30-40%
3. 40-50%
4. 50 – 60%

Answer: 4. 40-50%

Question 32. Which of the following gases may be used as a coolant?

1. CCl4
2. CFC
3. 03
4. None

Answer: 2. Cfc

Question 33. Which layer is damaged due to supersonic jets?

1. O3 layer
2. CO₂ layer
3. Co layer
4. None

Answer: 1. O3 layer

Question 34. The following gas causes depletion of 02 from hemoglobin

1. CO₂
2. Co
3. Ch4
4. No

Answer: 2. Co

Question 35. Which are the primary constituents of photochemical smog?

1. Hydrocarbons and CFCS
2. CO₂ & NO₂
3. NO₂ & hydrocarbons
4. SO₂ & Co.

Answer: 3. NO₂ & hydrocarbons

Question 36. The greenhouse effect corresponds to

1. Water pollution
2. SO₂ poisoning
3. Global warming
4. Production of green manure

Answer: 3. Global warming

Question 37. Excessive nutrients in the water body are also called

1. Bio-accumulation
2. Bio-stimulation
3. Bio-magnification
4. Eutrophication

Answer: 4. Eutrophication

Life Science Class 10 Chapter 5 Mcqs With Answers

Question 38. The nitrogen-fixing bacteria only become active

1. In the presence of phosphorus
2. In the presence of a high amount of nitrogen
3. In the absence of nitrogen
4. In low concentrations of nitrogen

Answer: 3. In the absence of nitrogen

Question 39. The following element is highest in seawater

1. Cl
2. F
3. P
4. C

Answer: 3. P

Question 40. Radiation may induce

1. Leukemia
2. Blood sugar
3. Heart attack
4. Osteoporosis

Answer: 1. Leukemia

Question 41. The following metal blocks hemoglobin biosynthesis

1. Co
2. Fe
3. Cu
4. Pb

Answer: 4. Pb

Question 42. Acid rain may include

1. Nitrous acid
2. Nitric acid
3. Sulphuric acid
4. All of them

Answer: 4. All of them

Question 44. Exposure to the following radioactive elements causes bone cancer & leukemia

1. Radium 226
2. Thorium232
3. Strontium90
4. Uranium236

Answer: 3. Strontium90

Life Science Class 10 Chapter 5 Mcqs With Answers

Question 45. The gradual increase of a pollutant along the food chain is also called

1. Bio-pollution
2. Bio-enhancement
3. Bio-remediation
4. Bio-magnification

Answer: 4. Bio-magnification

Question 46. A pollutant is any substance, chemical, or other factor, that changes

1. The natural balance of our environment
2. Natural geochemical cycles
3. The natural flora of our environment
4. Natural wildlife of our region

Answer: 1. The natural balance of our environment

Question 47. Which of the following is associated with biological amplification?

1. S04
2. Sewage
3. Ddt
4. All of these

Answer: 3. Ddt

Question 48. Non-biodegradable organic pollutants include

1. NPK
2. DDT
3. Nitric acid
4. Polyhydroxy butyrate

Answer: 2. Ddt

Question 49. The threshold of pain in human ears is

1. 50-60 Db
2. 30-40 Db
3. 160-180 Db
4. 120-140 Db

Answer: 4. 120-140 Db

Class 10 Life Science Multiple Choice Questions

Question 50. The component of living cells affected by the pollutant SO₂ is

1. All cell membrane system
2. Nucleus
3. Cell wall
4. Mitochondria

Answer: 1. All cell membrane system

Question 51. The following organism produces biodegradable plastic

1. Algae
2. Fungi
3. Bacteria
4. Virus

Answer: 3. Bacteria

Question 52. The pollutant responsible for the luxuriant growth of algae which forms water bloom is

1. Phosphates
2. DDT
3. H2s
4. Sulfates

Answer: 1. Phosphates

Question 53. The highest ddt deposition shall occur in

1. Phytoplanktons
2. Seagull
3. Crabs
4. Eels

Answer: 2. Seagull

Question 54. Drained sewage has a bod

1. More than that of clear water
2. Less than that of clear water
3. Equal to that of clear water
4. None of these

Answer: 1. More than that of clear water

Question 55. Who among the following are nature’s water cleaners?

1. Producers
2. Consumers
3. Symbionts
4. Decomposers & scavangers

Answer: 4. Decomposers & scavengers

WBBSE Chapter 5 Environment And Human Population MCQs

Question 56. Population is

1. Individuals in a family
2. Species in a community
3. Individuals in a species
4. Communities in an ecosystem

Answer: 1. Individuals in a species

Question 57. The rate of adding new ones produced in a specific period is termed as

1. Mortality
2. Popularity
3. Fatality
4. Natality

Answer: 4. Natality

Class 10 Life Science Multiple Choice Questions

Question 58. Resources are of two types- renewable resources and non-renewable resources.

1. Human
2. Economic
3. Geological
4. Natural

Answer: 4. Natural

Question 59. Is the destruction or clearing of forested lands, usually to expand agricultural land or for timber harvesting.

1. Forestation
2. Deforestation
3. Agriculture
4. Silting

Answer: 2. Deforestation

Question 60. Shrinking fields is one of the biggest challenges for making food available to the world population.

1. Agricultural
2. Wheat
3. Paddy
4. Both ‘2’ and ‘3’ correct

Answer: 1. Agricultural

Question 61. The human carrying capacity of the earth is between

1. 6 To 7 billion
2. 7 To 8 billion
3. 8 To 9 billion
4. 9 To 10 billion

Answer: 4. 9 To 10 billion

Question 62. Pollution has been linked to health problems like asthma and lung disease.

1. Air
2. Water
3. Noise
4. Thermal

Answer: 1. Air

Question 63. Zero population growth is characterized by

1. No growth in population recorded in the last decade
2. The very slow rate of growth of the population
3. Constant rate of growth of population
4. Stationary growth of population in one calendar year

Answer: 1. No growth in population recorded in the last decade

Class 10 Life Science Multiple Choice Questions

Question 64. Problem(s) due to ever-increasing population is/are

1. Depletion of natural resources
2. Deforestation
3. Increased pollution
4. All of these

Answer: 4. All of these

Question 65. Wetlands act as the biological of the landscape

1. Lungs
2. Liver
3. Heart
4. Kidneys

Answer: 4. Kidneys

Question 66. Lower rates, along with better management of land and water resources, are necessary to avert chronic food shortages.

1. Death
2. Birth
3. Growth
4. Respiration

Answer: 2. Birth

Question 67. Produce an exaggerated response of the immune system in which a specific antibody initiates the inflammatory response.

1. Mutagens
2. Nitrogen
3. Allergens
4. Hydrogen

Answer: 3. Allergens

Question 68. Bronchitis is a disease in which the mucus membrane in the lung’s bronchial passages becomes inflamed.

1. Circulatory
2. Respiratory
3. Excretory
4. Fungal

Answer: 2. Respiratory

Question 69. Is an uncontrolled growth of cells that disrupts body tissues and organs.

1. Tumor
2. Cancer
3. Apoptosis
4. Leukemia

Answer: 2. Cancer

Question 70. Experts agree that is the single biggest avoidable cause of cancer in the world.

1. Tobacco
2. Salt
3. Sugar
4. Radiation

Answer: 1. Tobacco

Class 10 Life Science Multiple Choice Questions

Question 71. The declining phase of a population is defined by

1. Mortality > natality
2. Natality > mortality
3. Mortality = natality
4. Natality = mortality = 0

Answer: 1. Mortality > natality

Question 72. An example of an indoor allergen is

1. Pollen
2. Dust mites
3. Stone dust
4. Smoke

Answer: 2. Dust mites

Question 73. Factor(s) responsible for bronchitis is/are

1. Smoking
2. Air pollution
3. Occupational irritants
4. All of these

Answer: 4. All of these

Question 74. An example of a disease caused due to air pollution is

1. Jaundice
2. Typhoid
3. Asthma
4. Amoebiasis

Answer: 3. Asthma

Question 75. A common symptom of asthma is

1. Fever
2. Breathing difficulty
3. Burning sensation
4. Loss of appetite

Answer: 2. Breathing difficulty

Life Science Chapter 5 Class 10 Mcq Solutions

Question 76. If the number of pollen grains, fungal spores, and dust particles increase suddenly in the air, decide which of the following problems may increase-

1. Tuberculosis
2. Asthma
3. Malaria
4. Dengu

Answer: 2. Asthma

WBBSE Chapter 5 Topic D Biodiversity And Conservation MCQs

Question 77. Biodiversity describes the diversity of

1. Plants
2. Animals
3. Life
4. Wildlife

Answer: 3. Life

Question 78. Is a unit of the biosphere.

1. Ecosystem
2. Species
3. Biodiversity
4. Ecology

Answer: 1. Ecosystem

Question 79. The worst dangerous threat to wildlife is by

1. Hunting
2. Habitat destruction
3. Overgrazing
4. Exotic species

Answer: 2. Habitat destruction

Question 80. The Equatorial region of the earth is rich in

1. Megadiversity
2. Biodiversity
3. Minerals
4. Rainfall

Answer: 2. Biodiversity

Life Science Chapter 5 Class 10 Mcq Solutions

Question 81. Sexual reproduction plays an important role in diversity.

1. Bio
2. Eco
3. Genetic
4. Mega

Answer: 3. Genetic

Question 82. A buffer zone of the biosphere reserve is where

1. Wild animals are absent here
2. No human activity is allowed
3. Several human activities freely occur
4. Limited human activity is permitted

Answer: 4. Limited human activity is permitted

Question 83. The gums and adhesives are produced from the tree

1. Acacia
2. Arabia
3. Ficus
4. Hibiscus

Answer: 1. Acacia

Question 84. Resins are a useful product of gymnospermous wood like

1. Fern
2. Rose
3. Pinus
4. Timber

Answer: 3. Pinus

Question 85. Apis indica produces

1. Honey
2. Wax
3. Alcohol
4. Both ‘1’ and ‘2’ correct

Answer: 4. Both ‘1’ and ‘2’ correct

Question 86. The incorrect match for tiger reserve amongst the following is

1. Buxa – West Bengal
2. Bandipur – Karnataka
3. Kanha – Madhya Pradesh
4. Palamau – Tamilnadu

Answer: 4. Palamau – tamilnadu

Question 87. Pearls are produced by

1. Sea
2. Oysters
3. Instruments
4. Marine animals

Answer: 2. Oysters

Life Science Chapter 5 Class 10 Mcq Solutions

Question 88. Globally, the number of biodiversity hotspots are

1. 4
2. 24
3. 35
4. 44

Answer: 3. 35

Question 89. In India, rhinoceros is found in the biodiversity hotspot of

1. Eastern Himalayas
2. Western ghats
3. Indo-Burma
4. Sundaland

Answer: 1. Eastern Himalayas

Question 90. Which is not a reason for the loss of biodiversity?

1. Over-exploitation
2. Over-eating
3. Pollution
4. Introduction of exotic species

Answer: 2. Over-eating

Question 91. Which is the first national park established in India?

1. Bandipur
2. Periyar
3. Kanha
4. Corbett

Answer: 4. Corbett

Question 92. Is considered a a world heritage site by UNESCO.

1. Kolkata
2. Delhi
3. Sundarbans
4. Himalayas

Answer: 3. Sundarbans

WBBSE Class 10 Life Science Chapter 5 MCQs

Question 93. The diversity of organisms sharing the same community or habitat is called

1. Alpha diversity
2. Beta diversity
3. Gamma diversity
4. None of these

Answer: 1. Alpha diversity

Question 94. Ex-situ conservation may be done in

1. Gene bank
2. National park
3. Zoo gardens
4. Sanctuary

Answer: 3. Zoo gardens

Question 95. The joint forest management system originated in

1. West Bengal
2. Bihar
3. Uttar Pradesh
4. Sikkim

Answer: 1. West Bengal

Question 96. Tiger conservation initiative in India was started in

1. 1743
2. 1973
3. 2005
4. 1947

Answer: 2. 1973

Question 97. In India, the red panda is found in

1. Bihar
2. Kerala
3. Karnataka
4. Sikkim

Answer: 4. Sikkim

Question 98. In India, lions are conserved in

1. Sundarban
2. Manas
3. Nandankanan
4. Gir forest

Answer: 4. Gir forest

Question 99. The first biosphere reserve established in India for conserving the gene pool of flora and fauna and the lifestyle of tribals is

1. Nilgiri biosphere reserve
2. Nanda Devi Biosphere Reserve
3. Thar biosphere reserve
4. Great Nicobar Biosphere reserve

Answer: 1. Nilgiri Biosphere reserve

WBBSE Class 10 Life Science Chapter 5 MCQs

Question 100. The ex-situ conservation of biodiversity is done through

1. National park
2. Biosphere reserve
3. Cryopreservation of gametes
4. Wildlife sanctuary

Answer: 3. Cryopreservation of gametes

Question 101. Environmental problems affecting the Sunderban ecosystem is/are due to

1. Pollution
2. Fresh water crisis
3. Urbanization
4. All of these

Answer: 4. All of these

Objective Type Questions MCQs

Question 102. One endangered species of the Eastern Himalayan biodiversity hotspot is

1. Lion-tailed macaque
2. Orangutan
3. Red Panda
4. Nilgiri tahr

Answer: 3. Red Panda

Question 103. An example of ex-situ conservation is

1. Sunderban Tiger Reserve
2. Corbet National Park
3. Nilgiri Biosphere Reserve
4. Cryopreservation

Answer: 4. Cryopreservation

Question 104. Gorumara, Corbett, Kulik, Nandadevi— choose the correct sequence from the following which is correct for the above forests in sequence

1. Biosphere Reserve, Sanctuary, National Park, National Park
2. National Park, National Park, Biosphere Reserve, Sanctuary
3. National Park, National Park, Sanctuary, Biosphere Reserve
4. Sanctuary, Biosphere Reserve, National Park, Sanctuary

Answer: 3. National Park, National Park, Sanctuary, Biosphere Reserve

WBBSE Chapter 4 Survival Strategies Adaptation MCQs

Question 1. Animal migration is an example of adaptation.

1. Behavioral
2. Structural
3. Functional
4. Physiological

Answer: 1. Behavioral

Question 2. Most species have reduced or lost true leaves, retaining only spines, which are highly modified leaves.

1. Sundari
2. Roses
3. Cacti
4. Plants

Answer: 3. Cacti

Read and Learn More WBBSE Class 10 Life Science Multiple Choice Questions

Question 3. The spines protect the cactus from

1. Flies
2. Birds
3. Sunlight
4. Predators

Answer: 4. Predators

WBBSE Class 10 Life Science Chapter 4 MCQs

Question 4. Over time, animals that are better adapted to their environment survive and

1. Feed
2. Breed
3. Evolve
4. All above are correct

Answer: 2. Breed

Question 5. Evolution is the result of

1. Ecology
2. Adaptation
3. Survival
4. Breeding

Answer: 2. Adaptation

Question 6. The swim bladder of a fish helps in

1. Respiration
2. Circulation
3. Buoyancy
4. Excretion

Answer: 3. Buoyancy

Question 7. In fish rete mirabile is found in

1. Swim bladder
2. Heart
3. Gill
4. None of the above

Answer: 1. Swim bladder

WBBSE Class 10 Life Science Chapter 4 MCQs

Question 8. The number of air sacs in birds is usually

1. 2
2. 5
3. 15
4. 9

Answer: 4. 9

Question 9. An example of a mangrove plant is

1. Mango
2. Garan
3. Ginger
4. Potato

Answer: 2. Garan

Question 10. In their hump camels store

1. Water
2. Fat
3. Energy
4. Food

Answer: 2. Fat

Question 11. One of the chimpanzees’ favorite foods is

1. Tomato
2. Potato
3. Nuts
4. Rice

Answer: 3. Nuts

WBBSE Class 10 Life Science Chapter 4 MCQs

Question 12. Did extensive study of the behavior of chimps.

1. Kohler
2. Darwin
3. Mendel
4. Newton

Answer: 1. Kohler

Question 13. The language of honey bees was decoded by

1. Kohler
2. Darwin
3. Mendel
4. Karl von Frisch

Answer: 4. Karl von frisch

Question 14. Honeybees indicate the direction of food sources by

1. Round dance
2. Waggle dance
3. Rock dance
4. Both ‘1’ and ‘2’ are correct

Answer: 2. Waggle dance

Question 15. The erythrocytes of camels are

1. Oval shaped
2. Nonnucleated
3. Capable of resisting osmotic variations
4. All of these

Answer: 4. All of these

Question 16. An example of behavioral adaptation is

1. Viviparous germination
2. Dancing of worker bees
3. Modification of leaf into the spine
4. Presence of swim bladder

Answer: 4. Presence of swim bladder

Question 17. Phyllocladeisfoundin

1. Xerophytes
2. Mesophytes
3. Halophytes
4. Hydrophytes

Answer: 1. Xerophytes

Life Science Class 10 MCQs Chapter 4 Solutions

Question 18. Which of the following adaptations enables the camel to live in deserts?

1. The hump stores water
2. The thin skin acts as a good heat insulator
3. Decreased GFR in kidney
4. Large nostrils help easy expiration of water vapor

Answer: 3. Decreased GFR in kidney

Question 19. Which of the following animals shows double respiration?

1. Toad
2. Rohu fish
3. Snakes
4. Pigeon

Answer: 4. Pigeon

WBBSE Class 10 Life Science Chapter 4 MCQs With Answers

Question 20. Physiologically dry soil contains

1. Adequate water
2. No water
3. A large amount of saline water
4. A small amount of water

Answer: 3. A large amount of saline water

WBBSE Chapter 4 Evolution MCQs

WBBSE Chapter 3 Topic A Heredity MCQs

Question 1. The mechanism of transmission of characters from the parental generation to the offspring is called as

1. Genetics
2. Heredity
3. Immunity
4. Eugenics

Answer: 4. Eugenics

Question 2. The idea of mixing parental characters with the offspring is known as

1. Inheritance
2. Particulate inheritance
3. Blending inheritance
4. Variation

Answer: 3. Blending inheritance

Question 3. Environmental variations are

1. Not inherited
2. Inherited
3. Congenital
4. Harmful

Answer: 1. Not inherited

Read and Learn More WBBSE Class 10 Life Science Multiple Choice Questions

Question 4. Somatic variations are

1. Acquired
2. Genetic
3. Inherited
4. Transmissible

Answer: 1. Acquired

Question 5. Variations are usually caused by

1. Evolution
2. Adaptation
3. Asexual reproduction
4. Mutation

Answer: 4. Mutation

Question 6. Which one of the following pairs is not contrasting?

1. Tall & dwarf
2. Axial & terminal
3. Yellow & green
4. Round & light

Answer: 4. Round & light

Wbbse Class 10 Life Science Mcqs

WBBSE Class 10 Life Science Chapter 3 MCQs With Answers

Question 7. Mendel performed experiments on

1. Fruit fly
2. Guineapig
3. Garden pea (pisum sativum)
4. Sweet pea (lathynus)

Answer: 3. Garden pea (Pisum sativum)

Question 8. Mendel’s concept gave birth to the theory of inheritance.

1. Particulate
2. Blending
3. Mendelian
4. Darwinian

Answer: 1. Particulate

Question 9. An organism with two copies of the same allele is

1. Heterozygous for that trait
2. Heterologous for that allele
3. Homozygous for that trait
4. Homologous for that allele

Answer: 3. Homozygous for that trait

Question 10. Mendelian laws of heredity are connected with

1. Segregation, dominance & independent assortment
2. Segregation, independent assortment & re-combination
3. Gene linkage, segregation & independent assortment
4. Gene linkage, dominance & independent assortment

Answer: 1. Segregation, dominance & independent assortment

Wbbse Class 10 Life Science Mcqs

Question 11. Gene mutations are also known as

1. Inversion
2. Point mutation
3. Deletion
4. Translocation

Answer: 2. Point mutation

Question 12. A diploid individual’s chromosome is

1. N
2. 2N
3. 3N
4. 2N+l

Answer: 4. 2N+l

Question 13. In humans, tongue-rolling characteristics are found in

1. A few
2. All
3. None
4. Most

Answer: 4. Most

Question 14. The attached ear lobe character in humans is

1. Non-genetic
2. Dominant
3. Recessive
4. Very useful

Answer: 3. Recessive

Question 15. An alternative form of a gene, located at a specific position on a specific chromosome is called

1. Locus
2. Trait
3. Variety
4. Allele

Answer: 4. Variety

Question 16. The specific location of a gene on a chromosome is called its

1. Locus
2. Allele
3. Trait
4. Variety

Answer: 1. Locus

Wbbse Class 10 Life Science Chapter 3 Solutions 

Question 17. Mendelian factors are now called

1. Alleles
2. Genes
3. Mutants
4. Loci

Answer: 2. Genes

Question 18. A genetic cross dealing with two character differences is called

1. Monohybrid cross
2. Dihybrid cross
3. Trihybrid cross
4. None of the above

Answer: 2. Dihybrid cross

Question 19. The genotype ‘as’ is

1. Allozygous
2. Heterozygous
3. Homozygous
4. Hemizygous

Answer: 2. Heterozygous

Question 20. The character that is expressed in the heterozygous condition is

1. Allele
2. Hybrid
3. Dominant
4. Recessive

Answer: 3. Dominant

WBBSE Life Science Chapter 3 Heredity Solved MCQs

Question 21. The outer expression of a character is its

1. Phenotype
2. Genotype
3. Trait
4. Allele

Answer: 1. Phenotype

Question 22. Pea plants normally perform

1. Asexual reproduction
2. Vegetative reproduction
3. Cross-fertilization
4. Self-fertilization

Answer: 4. Self-fertilization

Wbbse Class 10 Life Science Chapter 3 Solutions 

Question 23. In case of incomplete dominance in the F₂ generation

1. The genotypic ratio = 3:1
2. The phenotypic ratio = 3:1
3. The genotypic ratio = phenotypic ratio
4. The genotypic ratio phenotypic ratio

Answer: 3. The genotypic ratio = phenotypic ratio

Question 24. The F₂ phenotypic ratio of the Mendelian dihybrid cross is

1. 1:1:1:1
2. 9:3:3:L
3. 3:1
4. 1:2:1

Answer: 3. 3:1

Question 25. The case of incomplete dominance is observed in

1. Pisum sativum
2. Drosophila melanogaster
3. Mirabilis jalapa
4. All of the above

Answer: 3. Mirabilis jalapa

Question 26. Human males are

1. Homogametic
2. Heterogametic
3. Homozygous
4. Heterozygous

Answer: 2. Heterogametic

Question 27. Sry is present in human

1. Y chromosome
2. X chromosome
3. Autosome
4. Both ‘1’ and ‘2’ correct

Answer: 1. Y chromosome

Class 10 Life Science Mcq With Answers

Question 28. In men, sperm contain autosomes and

1. Both the x and y chromosomes
2. Either the x or y chromosome
3. Only y chromosome
4. Only x chromosome

Answer: 2. Either x or y chromosome

Question 29. If a homozygous dominant red-flowered plant is crossed with a homozygous recessive white-flowered plant, the offspring will be

1. Half red flowered
2. Half white flowered
3. All red flowered
4. All pink flowered

Answer: 3. All red flowered

Question 28. Independent assortment can be proved by

1. Back cross
2. Test cross
3. Monohybrid cross
4. Dihybrid cross

Answer: 2. Test cross

Question 29. A cross between black and white birds results in all blue in the F₁ generation. Then the cross between bluebirds in the F₂ generation would result in

1. 3 Blue: 1 white
2. 1 Blue: 1 black: 1 white
3. 9 Blue : 3 black : 3 white
4. 1 Black: 1 white: 2 blue

Answer: 3. 9 Blue : 3 black : 3 white

Question 30. The four daughter cells derived from single meiosis differ from each other due to

1. The difference is chromosome number
2. Crossing over only
3. Independent assortment of chromosomes only
4. Crossing over and independent assortment of chromosomes

Answer: 4. Crossing over and independent assortment of chromosomes

Question 31. The number of distinct traits selected by Mendel in garden peas for his breeding experiments are

1. 7
2. 2
3. 4
4. 8

Answer: 4. 8

Class 10 Life Science Mcq With Answers

Question 32. When two hybrids undergo crossing in a monohybrid cross, then the percentage of recessive is

1. 50
2. 25
3. 100
4. 75

Answer: 4. 75

Question 33. If a couple has 5 daughters, the percentage probability of the other child being a girl is

1. 25
2. 100
3. 75
4. 50

Answer: 1. 25

Objective Type Questions MCQs

Question 1. Which of the following traits in the pea plant is recessive?

1. Wrinkled seed
2. Yellow colored seed
3. Purple colored flower
4. Axial flower

Answer: wrinkled seed.

Question 2. How many types of gametes are formed from pea plants having the genotype yy rr?

1. 1
2. 4
3. 2
4. 3

Answer: 4.

Question 3. Which one of the following is the genotypic ratio at the F₂ generation of Mendel’s monohybrid cross?

1. 1:2:1
2. 3:1
3. 9:3:3:1
4. 2:1:2

Answer: 1:2:1

Question 4. What would be the phenotypic ratio in the F₂ generation of a monohybrid cross in case of incomplete dominance?

1. 3:1
2. 2:1:1
3. 9:3:3:L
4. 1:2:1

Answer: 1:2:1

Question 5. Identify which of the following is a dominant trait

1. Length of the stem – dwarf
2. The shape of the seed – wrinkled
3. Color of the cotyledon – yellow
4. Color of the flower – white

Answer: the color of the cotyledon – yellow

Life Science Class 10 Wbbse Multiple Choice

Question  6. Assess how many types of gametes are produced from the pea plant having the genotype rryy —

1. One type
2. Four types
3. Two types
4. Three types

Answer: one type

Question 7. Select which of the following two genotypes are responsible for the expression of the phenotype wrinkled yellow in Pea plant

1. RRYY and rryy
2. RRYy and RrYy
3. RRyy and Rryy
4. rrYY and rrYy

Answer: rrYY and rrYy

Life Science Class 10 Wbbse Multiple Choice

WBBSE Chapter 3 Some Common Genetic Diseases MCQs

Question 1. Genetic Disorders Are

1. Harmless
2. Inherited
3. Contagious
4. Lethal

Answer: 2. Inherited

Question 2. Haemophilia Is Related To

1. Blood
2. Bone
3. Nerve
4. Kidney

Answer: 1. Blood

Question 3. Color Blindness Is

1. Sex-Linked Dominant
2. Autosomal Dominant
3. Sex-Linked Recessive
4. Autosomal Recessive

Answer: 3. Sex-Linked Recessive

Question 4. Thalassemia Is

1. Infectious
2. Curable
3. Environmental
4. Genetic

Answer: 4. Genetic

Wbbse Class 10 Life Science Mcqs

Question 5. Haemophilia B Is Also Known As

1. Anaemia
2. Thalassemia
3. Christmas Disease
4. Factor Viii Deficiency

Answer: 3. Christmas Disease

Question 6. The Most Common Type Of Colour Blindness Is

1. Red-Green
2. Blue-Yellow
3. Blue-Green
4. None Above

Answer: 1. Red-Green

Question 7. The Chromosome Carrying Sex-Linked Gene Is

1. Autosomes
2. Y Chromosome
3. X Chromosome
4. None Of These

Answer: 3. X Chromosome

Question 8. Holandric Genes Are Ones Situated On

1. X Chromosome
2. Y Chromosome
3. Both A & B
4. Autosomes

Answer: 2. Y Chromosome

Question 9. A Man Known To Be A Victim Of Haemophilia Mar-Ries A Normal Woman Whose Father Was Known To Be Bleeder; Then It May Be Expected That

1. All Their Children Would Be Bleeders
2. Half Of Their Sons Would Be Bleeders
3. One-fourth of Their Children Would Be Bleed-Ers
4. All Would Be Normal

Answer: 2. Half Of Their Sons Would Be Bleeders

Wbbse Class 10 Life Science Chapter 3 Solutions

Question 10. Red Green Colour Blindness Is Also Known As

1. Protanopia
2. Deuteranopia
3. Phobia
4. Both 1&2

Answer: 4. Both 1&2

Question 11. A Colourblind Man, Both Of Whose Parents Had Normal Vision And Whose Paternal And Maternal Grandparents Had Normal Vision, Probably Inher-Ited The Gene For Colourblindness From His

1. Maternal Or Paternal Grandmother
2. Maternal Or Paternal Grand Father
3. Mother
4. Father

Answer: 3. Mother

Question 12. Daltonism Can Occur In Females Due To

1. Homozygous Condition In Which Both X chromosomes Are Mutated At The Same Locus
2. Heterozygous Condition Of X Chromosome
3. Homozygous Condition Of Y Chromosome
4. None Of These

Answer: 1. Homozygous Condition In Which Both X chromosomes Are Mutated At The Same Locus

Question 13. The Daughters Born To a Haemophilic Father And Normal Mother Could Be

1. Normal
2. Carrier
3. Haemophilic
4. All Of These

Answer: 2. Carrier

Question 14. Color Blindness Is Related to a Defect In

1. Rods
2. Cones
3. Eye Muscles
4. Cornea

Answer: 2. Cones

Question 15. Haemophilia Is A Genetic Disorder In Which

1. Blood Fails To Coagulate After An Injury
2. There Is Delayed Coagulation Of the Blood
3. Blood Clots In Blood Vessels
4. Blood Cell Count Falls

Answer: 2. There Is Delayed Coagulation Of Blood

Question 16. A Cross Between A Normal Homozygous Woman And A Haemophilic Man Will Result In How Many Sons Having The Above Disease?

1. 0%
2. 50%
3. 75%
4. 100%

Answer: 1. 0%

Question 17. If You Are Missing One Of The Four Alpha Chain Genes, Then You Are Suffering From Which Type Of Alpha Thalassemia?

1. Silent Carrier
2. Alpha Thalassemia Minor
3. Alpha Thalassemia Major
4. Haemoglobin H Disease

Answer: 1. Silent Carrier

Wbbse Class 10 Life Science Chapter 3 Solutions

Question 18. Which Type Of Thalassemia Is Also Called Cooley’s Anaemia?

1. Alpha Thalassemia Major
2. Beta Thalassemia Major
3. Beta Thalassemia Minor
4. Hydrops Fetalis

Answer: 2. Beta Thalassemia Major

Question 19. What Are The Factors That Increase The Risk Of Thalassemia Disease?

1. Family History
2. Certain Ancestry
3. Both 1 & 2
4. Superstition

Answer: 3. Both 1 & 2

Question 20. A Good Genetic Counsellor Should Have Skills Of

1. Genetics And Hereditary Disease
2. Active Listening, Comprehension & Complex Problem Solving
3. Social Perceptiveness
4. All Of These

Answer: 4. All Of These

WBBSE Chapter 2 Cell Division And Cell Cycle MCQs

1. Lamarck
2. Fleming
3. Virchow
4. Schleiden

1. The cell’s DNA replicates during G1
2. It consists of mitosis and interphase
3. A cell can remain in g1 for weeks or much longer
4. Most proteins are formed throughout all subphases of interphase

1. Nuclei
2. Chromosomes
3. Cell membranes
4. Vacuoles

Answer: 2. Chromosomes

Read and Learn More WBBSE Class 10 Life Science Multiple Choice Questions

Question 4. The main function of a gene is to

1. Regulate cellular respiration
2. Synthesize proteins & enzymes and guide cell function
3. Assist the metabolism of fat
4. All the above

Answer: 4. All the above

WBBSE Class 10 Life Science Chapter 2 MCQs

Question 5. Which is not a nitrogenous base of DNA?

1. Adenine
2. Guanine (g)
3. Cytosine
4. Uracil(u)

Answer: 4. Uracil(u)

Question 6. Are long thread-like structures made of a DNA molecule and protein.

1. Nuclei
2. Chromosomes
3. Genes
4. RNAs

Answer: 2. Chromosomes

Question 7. Each of our body cells contains pairs of chromosomes.

1. 23
2. 46
3. 11
4. None of the above

Answer: 1. 23

Question 8. The sex chromosomes of human males are

1. XX
2. XY
3. YY
4. ZZ

Answer: 2. XY

Question 9. Pictures of chromosomes arranged in pairs are known as

1. Histogram
2. Chronogram
3. Cardiogram
4. Karyogram

Answer: 4. Karyogram

WBBSE Class 10 Life Science Chapter 2 MCQs

Question 10. Cells with a haploid number of chromosomes are produced by

1. Mitosis
2. Amitosis
3. Meiosis
4. Cell division

Answer: 3. Meiosis

WBBSE Class 10 Life Science Chapter 2 MCQs With Answers

Question 11. The twenty-third pair of chromosomes in man is known as

1. Heterosome
2. Chromatid
3. Autosome
4. Gene

Answer: 1. Heterosome

Question 12. At mitotic metaphase, each chromosome consists of two symmetrical structures called

1. Centrioles
2. Genes
3. Spindle fibers
4. Chromatids

Answer: 4. Chromatids

Question 13. Are v-shaped chromosomes in which the centromere lies in the middle of the chromosome so that the two arms are almost equal.

1. Telocentric
2. Acrocentric
3. Metacentric
4. Sub-metacentric

Answer: 3. Metacentric

Question 14. Secondary constrictions are also called

1. DNA
2. NOR
3. RNA
4. Centromeres

Answer: 2. NOR

WBBSE Class 10 Life Science Chapter 2 MCQs

Question 15. Different types of chromosomes can be recognized by the position of the following separating the two arms

1. Telomeres
2. Micromeres
3. Macromeres
4. Centromeres

Answer: 4. Centromeres

Question 16. Chromatin is composed of DNA, RNA and

1. Gene
2. Nucleic acid
3. Protein
4. All of the above

Answer: 3. Protein

Question 17. Deoxyribose sugar is

1. Pentose
2. Hexose
3. Triose
4. None of the above

Answer: 1. Pentose

Question 18. Chromosomes contain proteins.

1. Histone
2. Peptone
3. Chromatin
4. Ribose

Answer: 1. Histone

Question 19. Regions of the chromosomes containing genes are

1. Euchromatin
2. Chromatin
3. Heterochromatin
4. Centromere

Answer: 1. Euchromatin

Continuity Of Life MCQs For Class 10 Board Exams

Question 20. Cytokinesis refers to

1. Division of nuclei
2. Division of cytoplasm
3. Division of chromosomes
4. Completion of cell division

Answer: 2. Division of cytoplasm

Question 21. In sexual reproduction

1. The chromosome number is reduced during mitosis
2. The zygote is usually haploid
3. Gametes are usually haploid
4. Gametes are usually diploid

Answer: 3. Gametes are usually haploid

Question 22. Microtubules are thick, strong spirals of thousands of subunits known as\

1. Proteins
2. Fibers
3. Spindles
4. Tubulin

Answer: 4. Tubulin

Question 23. Mitochondrial production is also vital for cell division.

1. DNA
2. ATP
3. BMR
4. ADP

Answer: 2. ATP

Question 24. Is also known as reductional cell division.

1. Mitosis
2. Meiosis
3. Amitosis
4. Binary fission

Answer: 2. Meiosis

Question 25. Is the time when a cell will leave the cycle and quit dividing.

1. G₀
2. G₁
3. G₂
4. S

Answer: 1. G₀

Class 10 Life Science Chapter 2 Multiple Choice

Question 26. The chromosome set is

1. The same for all organisms
2. Varies with the age of the organisms
3. Same for all the plants
4. Constant for a given species

Answer: 4. Constant for a given species

Question 27. Mitotic nuclear division is called

1. Cytokinesis
2. Akinesis
3. Karyokinesis
4. Ookinesis

Answer: 3. Karyokinesis

Question 28. Karyokinesis does not include

1. Prophase
2. Metaphase
3. Anaphase
4. Cytokinesis

Answer: 4. Cytokinesis

Question 29. Chromosomes are arranged in the equator of the cell during

1. Prophase
2. Metaphase
3. Anaphase
4. Telophase

Answer: 2. Metaphase

Question 30. Crossing over occurs during

1. Meiosis I
2. Meiosis II
3. Amitosis
4. Mitosis

Answer: 1. Meiosis I

MCQs For Continuity Of Life WBBSE

Question 31. Which of the following statements is true?

1. A diploid cell produces four diploid cells in meiosis
2. The chromosome number is reduced to half in daughter cells in mitosis
3. The chromosome number remains the same as the parent cell in meiosis
4. The daughter cell receives both maternal and paternal chromosomes of the homologous pair in meiosis

Answer: 4. The daughter cell receives both maternal and paternal chromosomes of the homologous pair in meiosis

Question 32. During mitosis, each chromosome at the beginning of the prophase is

1. Single-threaded
2. Paired structure
3. Four threaded
4. None of these

Answer: 2. Paired structure

Question 33. In mitotic cell division the

1. The amount of DNA in the daughter cells will be equal to the parent cell
2. The size will be half of the parent cell
3. DNA will be double the parent cell
4. Both (1) & (2)

Answer: 4. Both (1) & (2)

Question 34. Which of the following is true for mitosis?

1. No chiasma formation
2. No crossing over
3. Prophase has no substage
4. All of these

Answer: 4. All of these

West Bengal Board Class 10 Life Science MCQs

Question 35. Identify the pair of nucleotides that are joined by hydrogen bonds in double-stranded DNA.

1. AG & CG
2. TG & CG
3. AT & CG
4. At & TG

Answer: 3. AT & CG

Question 36. Chromosomes do not have a well-defined structure in

1. Interphase
2. Prophase
3. Metaphase
4. Telophase

Answer: 1. Interphase

Question 37. Crossing over is the

1. Exchange of genetic material
2. Deletion of chromosomes
3. Linkage of chromosomes
4. Inversion of chromosomes

Answer: 1. Exchange of genetic material

Question 38. If the number of chromosomes in most body cells of a mammal is 40, the cells in the seminiferous tubule will have

1. 40 Chromosomes
2. 23 Chromosomes
3. 20 Chromosomes
4. 11 Chromosomes

Answer: 3. 20 Chromosomes

Question 39. Chromosomes which have a definite role in sex determination are

1. Lysosomes
2. Allosomes
3. Giant chromosomes
4. Autosomes

Answer: 2. Allosomes

Question 40. Crossing over occurs between

1. Non-sister chromatids of homologous chromosomes
2. Sister chromatids of homologous chromosomes
3. Any chromosomes
4. Sister chromatids of non-homologous chromosomes

Answer: 1. Non-sister chromatids of homologous chromosomes

Question 41. Which of the following is true?

1. Meiosis maintains a constant number of chromosomes in an organism
2. Meiosis provides an opportunity for the exchange of genes
3. Meiosis causes genetic variations among the species
4. All of these

Answer: 4. All of these

Question 42. The characteristic structure called phragmoplast formed during plant cell division is the precursor of

1. Chloroplast
2. Phragmosomes
3. Cell wall
4. Vacuoles

Answer: 3. Cell wall

Question 43. There can not be mitotic division in one of the following cells. Identify the cell

1. Bone cell
2. Sperm
3. Zygote
4. New red blood cells

Answer: 2. Sperm

Question 44. The complete process of meiosis involves

1. One cytoplasmic division with only one chromosome reduction
2. Two cytoplasmic divisions with one reduction of chromosomes
3. Two cytoplasmic divisions with two reductions of chromosomes
4. One cytoplasmic division with two reductions of chromosomes

Answer: 2. Two cytoplasmic divisions with one reduction of chromosomes

Question 45. Amitosis is

1. Division involving the formation of chromosome bridges
2. Division involving spindle formation
3. Division in which chromosomes are unequally distributed
4. Cleavage of the nucleus without recognizable chromosome distribution

Answer: 4. Cleavage of the nucleus without recognizable chromosome distribution

West Bengal Board Class 10 Life Science MCQs

Chapter 2 Topic B Reproduction Review Questions MCQs

Question 46. Bulbil is a modification of

1. Stem
2. Root
3. Leaf
4. Bud

Answer: 4. Bud

Question 47. The animal that reproduces by multiple fission is

1. Amoeba
2. Paramoecium
3. Euglena
4. All of them

Answer: 4. All of them

Question 48. The organism that shows fragmentation is

1. Rhizobium
2. Glocecapsa
3. Oscillatoria
4. Bacillus

Answer: 3. Oscillatoria

Question 49. The most developed artificial propagation method is

1. Grafting
2. Gootee
3. Layering
4. Budding

Answer: 1. Grafting

Question 50. The following animal may develop by parthenogenesis:

1. Queen bee
2. Worker bee
3. Drones
4. Butterfly

Answer: 2. Worker bee

Question 51. The unit of asexual reproduction is

1. Bud
2. Gamate
3. Spore
4. Callus

Answer: 3. Spore

Class 10 Life Science Chapter 2 Important MCQs

Question 52. Which one is related to sexual reproduction?

1. Amitosis
2. Mitosis
3. Budding
4. Meiosis

Answer: 4. Meiosis

Question 53. Production of fruit without fertilization is known as

1. Parthenogenesis
2. Paedogenesis
3. Parthenocarpy
4. Neoteny

Answer: 3. Parthenocarpy

Question 54. Which type of reproduction paves the pathway of evolution?

1. Vegetative reproduction
2. Asexual reproduction
3. Sexual reproduction
4. Parthenogenesis

Answer: 3. Sexual reproduction

Question 55. Which one is not formed due to sexual reproduction?

1. Zoospore
2. Zygospore
3. Oospore
4. Zygote

Answer: 1. Zoospore

Question 56. Yeast reproduces by

1. Fission
2. Budding
3. Both
4. None

Answer: 3. Both

Question 57. The first diploid cell resulting from sexual reproduction is

1. Embryo
2. Zygote
3. Spore mother cell
4. Larva

Answer: 2. Zygote

Question 58. Which one is not an underground stem?

1. Rhizome
2. Tuber
3. Bulb
4. Runner

Answer: 4. Runner

Question 59. Gametes are

1. Haploid
2. Diploid
3. Either haploid or diploid
4. None of them

Answer: 1. Haploid

Question 60. The propagating organ for ginger is

1. Runner
2. Tuber
3. Rhizome
4. Corm

Answer: 3. Rhizome

Class 10 Life Science Chapter 2 Important MCQs

Question 61. The propagating organ for onion is

1. Offset
2. Sucker
3. Bullb
4. Bulbil

Answer: 3. Bullb

Question 62. The gametes that look identical are called

1. Isogamete
2. Anisogamete
3. Oogamete
4. Heterogamete.

Answer: 1. Isogamete

Question 63. The following organism may exhibit spore

1. Amoeba
2. Rhizopus
3. Paramoecium
4. Plasmodium

Answer: 2. Rhizopus

Question 64. The animal that does not show regeneration is

1. Planaria
2. Hydra
3. Ascaris
4. Starfish

Answer: 3. Ascaris

Question 65. Which one of the following methods is not a method of artificial vegetative propagation?

1. Grafting
2. Budding
3. Hybridization
4. Layering

Answer: 3. Hybridization

Question 66. The vegetative method of reproduction adopted for plants with longer seed dormancy is

1. Cutting
2. Layering
3. Budding
4. Grafting

Answer: 4. Grafting

Question 67. The part of the plant which remains close to the ground is

1. Stock
2. Scion
3. Sucker
4. Offset

Answer: 1. Stock

Question 68. Which one of the following reproduces with the help of a leaf?

1. Pteris
2. Oxalis
3. Bryophyllum
4. Banana

Answer: 3. Bryophyllum

Question 69. Which method results in the mixing of features of two separate plants?

1. Layering
2. Grafting
3. Gootee
4. Budding

Answer: 2. Grafting

Question 70. Which one of the following is true for vegetative reproduction?

1. It is both natural and artificial, producing genetically identical plants
2. It is natural
3. It is artificial
4. It is both natural and artificial, producing genetically different plants.

Answer: 4. It is both natural and artificial, producing genetically different plants.

Question 71. The following plant reproduces by vegetative means

1. Banana
2. Oxalis
3. Water hyacinth
4. All of them

Answer: 4. All of them

Question 72. The following hormone induces rooting after cutting

1. IBA
2. IAA
3. GA3
4. Ethylene

Answer: 2. IAA

Class 10 Life Science Chapter 2 Important MCQs

Question 73. The following plant reproduces by root bud

1. Dahlia
2. Sweet potato
3. Both
4. Water hyacinth

Answer: 3. Both

Question 74. Which one of the following processes is related to root induction

1. Cutting
2. Micropropagation
3. Grafting
4. Budding

Answer: 1. Cutting

Question 75. In grafting, the plant part that joins with the stock is called

1. Scion
2. Bud graft
3. Transplant
4. Shoot

Answer: 1. Scion

Question 76. The following plant reproduces with a fleshy root

1. Amanda
2. Dahlia
3. Asparagus
4. All of them

Answer: 4. All of them

Question 77. Potato tuber reproduces with

1. Apical bud
2. Axillary bud
3. Both
4. None

Answer: 3. Both

Question 78. The plant that reproduces by rhizome is

1. Ginger
2. Lotus
3. Turmeric
4. All of them

Answer: 4. All of them

Question 79. The plant that reproduces with Gootee is

1. Mango
2. Guava
3. Lemon
4. All of them

Answer: 4. All of them

Question 80. The animal that reproduces by conjugation is

1. Paramoecium
2. Tapeworm
3. Roundworm
4. Earthworm

Answer: 1. Paramoecium

Question 36. The largest plant production method of vegetative reproduction is

1. Stem cuttings
2. Root cuttings
3. Micropropagation
4. Grafting

Answer: 3. Micropropagation

Life Science Chapter 2 MCQs with Answers Class 10

Question 37. A totipotent cell means

1. An undifferentiated cell capable of developing into a system or entire plant
2. An undifferentiated cell capable of developing into a complete embryo
3. An undifferentiated cell capable of developing into an organ
4. A cell that cannot differentiate into an organ or system

Answer: 1. An undifferentiated cell capable of developing into a system or entire plant

Question 38. Which of the following plant cells show totipotency?

1. Sieve tubes
2. Xylem vessels
3. Meristem
4. Cork cells

Answer: 3. Meristem

Question 39. Asexually produced organisms inheriting all the characteristics of the parent are

1. Offspring
2. Clone
3. Variety
4. Hybrid

Answer: 2. Clone

Question 40. Which is connected to asexual reproduction?

1. Gemmules
2. Gametes
3. Gonads
4. Genitalia

Answer: 1. Gemmules

Question 41. Any cell, tissue, or organ removed from a plant for culturing is called

1. Stock
2. Scion
3. Explant
4. Embryoid

Answer: 2. Scion

Question 42. When two different individuals participate in vegetative reproduction, it is called

1. Cutting
2. Grafting
3. Pollination
4. Layering

Answer: 2. Grafting

Question 43. Parthenogenesis is

1. Development of embryo without fertilization
2. Development of fruit without fertilization
3. Development of fruit without hormone
4. Development of embryo from an egg without fertilization

Answer: 3. Development of fruit without hormone

Question 44. In which of the following modes of asexual reproduction, a mother cell produces innumerable unicellular uninucleate offspring?

1. Fragmentation
2. Multiple fission
3. Sporulation
4. Parthenogenesis

Answer: 2. Multiple fission

Question 45. In oogamy, fertilization involves

1. A large nonmotile female gamete and a small motile male gamete
2. A large nonmotile female gamete and a small nonmotile male gamete
3. A large motile female gamete and a small nonmotile male gamete
4. A small nonmotile female gamete and a large motile male gamete

Answer: 1. A large nonmotile female gamete and a small nonmotile male gamete

Question 46. Why is asexual reproduction sometimes disadvantageous?

1. It allows animals to produce many offspring quickly
2. It produces a genetically uniform population
3. It allows sedentary animals to produce offspring without mates
4. It saves time and energy in gamete formation

Answer: 2. It allows sedentary animals to produce offspring without mates

Life Science Chapter 2 MCQs with Answers Class 10

Question 47. When an animal is cut into pieces, each piece is found to grow into a complex organism. Name the process.

1. Budding
2. Fragmentation
3. Spore formation
4. Regeneration

Answer: 4. Regeneration

Question 48. In grafting contact is made between

1. Phloem
2. Xylem
3. Cambium
4. Flower

Answer: 3. Cambium

Question 49. Which of the following statements supports the view that elaborates sexual reproductive processes appeared much later in the organic evolution?

1. Lower groups of organisms have a simpler body design
2. Asexual reproduction is common in lower groups.
3. Asexual reproduction is common in higher groups of organisms.
4. The high incidence of sexual reproduction in angiosperms and vertebrates.

Choose the correct answer from the given options:

1. 1&3
2. 1&2
3. 2&4
4. 2&3

Answer: 3. 2&4

Question 50. There are various types of reproduction. The type of reproduction adopted by an organism depends on

1. The morphology of the organism
2. The habitat of the organism
3. The morphology and physiology of the organism
4. The habitat, physiology, and genetic makeup of the organism

Answer: 4. The habitat, physiology, and genetic makeup of the organism

Chapter 2 Topic C Sexual Reproduction In Flowering Plants Review Questions MCQs

1. 2
2. 3
3. 4
4. 6

1. Protection
2. Photosynthesis
3. Both
4. None

1. Autogamy
2. Allogamy
3. Geitonogamy
4. Xenogamy

1. Absence of Calyx and corolla
2. Absence of androecium
3. Absence of gynoecium
4. Both 2 & 3

1. Calyx
2. Corolla
3. Androecium
4. Gynoecium

1. Calyx
2. Corolla
3. Androecium
4. Gynoecium

1. Calyx
2. Corolla
3. Androecium
4. Gynoecium

1. Incomplete
2. Unisexual
3. Intersexual
4. Neuter

1. Calyx
2. Corolla
3. Androecium
4. Gynoecium

1. Secretion of nectar
2. Pollination
3. To increase the beauty of the plant
4. Reproduction

1. Rice
2. Mango
3. Orchid
4. Akanda

1. Magnolia
2. Sagittaria
3. Hydrilla
4. Akanda

1. Ovule
2. Embryo
3. Seed
4. Fruit

1. Simul
2. Paddy
3. Kadam
4. Neem

1. Corolla & carpel
2. Calyx & corolla
3. Calyx and gynoecium
4. Androecium & gynoecium

1. Corolla
2. Style
3. Stigma
4. Ovary

1. Cleistogamy
2. Homogamy
3. Herkogamy
4. All of these

1. Superior progeny
2. Weaker progeny
3. Formation of male offspring
4. Formation of seeds

1. 1
2. 2
3. 3
4. 4

1. N
2. 2N
3. 3N
4. 4N

1. Heterogamous
2. Homogamous
3. Dichogamous
4. Cleistogamous

1. Air
2. Water
3. Insects
4. Birds

1. Moths
2. Bees
3. Beetles
4. Butterflies

1. Pollination by birds
2. Pollination by insects
3. Pollination by wind
4. Pollination by water

1. Fusion of male gamete with antipodal cell
2. Fusion of male gamete with secondary nucleus
3. Fusion of male gamete with the egg cell
4. Fusion of one male gamete with the egg cell and another with a secondary nucleus

WBBSE Chapter 1 Sensitivity And Response In Plants Multiple Choice Questions

Question 1. The ability of a plant to react to external stimuli is

1. Movement
2. Locomotion
3. Evolution
4. Irritability

Answer: 4. Irritability

Question 2. The instrument with which sir. c. Bose proved the phenomenon of reacting to stimulus is

1. Seismograph
2. Crescograph
3. Lithograph
4. Sonometer

Answer: 2. Crescograph

Read and Learn More WBBSE Class 10 Life Science Multiple Choice Questions

Question 3. The movement of variation is controlled by

1. Diffusion
2. Osmosis
3. Transpiration
4. None of these

Answer: 2. Osmosis

Question 4. The following plant is capable of locomotion

1. Mimosa
2. Desmodium
3. Gourd
4. Volvox

Answer: 4. Volvox

WBBSE Class 10 Life Science Chapter 1 MCQs

Question 5. The movement offers antherozoid to archegonium is called

1. Photonasty
2. Chemonasty
3. Nyctinasty
4. Sismonasty

Answer: 2. Chemonasty

Question 6. The plant movement that is controlled by auxin is

1. Tactic
2. Nastic
3. Tropic
4. Movement of variation

Answer: 2. Nastic

Question 7. The movement of Chlamydomonas towards light is called

1. Phototaxis
2. Phototropism
3. Photonastism
4. Chemotaxis

Answer: 3. Photonastism

Question 8. The tactic movement controlled by an electric current is

1. Hydrotactic
2. Thermotactic
3. Rheotactic
4. Galvanotactic

Answer: 4. Galvanotactic

Question 9. The movement of plant roots against gravity is known as

1. Positively geotropic
2. Aerotropic
3. Positively hydrotropic
4. Positively rheotropic

Answer: 2. Aerotropic

WBBSE Class 10 Life Science Chapter 1 MCQs

Question 10. When a plant stem grows opposite to water, it is called

1. Positively phototropic
2. Negatively phototropic
3. Positively geotropic
4. Negatively geotropic

Answer: 2. Negatively phototropic

Question 11. The movement of drosera lea felt towards insects is called

1. Photonasty
2. Chemonasty
3. Nyctinasty
4. Seismonasty

Answer: 2. Chemonasty

Question 12. The plant movement depending on the intensity of the stimulus is

1. Tactic
2. Tropic
3. Nastic
4. Variation

Answer: 2. Tropic

Question 13. The nastic movement that is controlled by both light and temperature is

1. Photonasty
2. Thermonasty
3. Nyctinasty
4. Seismonasty

Answer: 3. Nyctinasty

Question 14. The movement of plants involving actual displacement is

1. Tropism
2. Taxes
3. Nastism
4. Amoeboid movement

Answer: 2. Taxism

Question 15. The movement of curvature whose direction is controlled by the direction of the stimulus is

1. Tactic
2. Tropic
3. Nastic
4. None of the above

Answer: 2. Tropic

WBBSE Class 10 Life Science Chapter 1 MCQs

WBBSE Class 10 Life Science Chapter 1 MCQs With Answers

Question 16. In comparison to geotropism, which one is a greater force?

1. Chemotropism
2. Hydrotropism
3. Thigmotropism
4. Phototropism

Answer: 2. Hydrotropism

Question 17. The following organism shows amoeboid movement

1. Myxamoeba
2. Fungi
3. Algae
4. Myxomycetes

Answer: 2. Fungi

Question 18. The movement that both Chlamys Doonas and Volvox exhibit is

1. Phototropism
2. Geotropism
3. Phototaxis
4. Chemotaxis

Answer: 3. Phototaxis

Question 19. The movement involved in the opening of the tulip flower is

1. Nyctinasty
2. Photonasty
3. Thigmonasty
4. Thermonasty

Answer: 4. Thermonasty

Class 10 Life Science Chapter 1 Multiple Choice

Question 20. The following plant movement is due to the growth

1. Nutation
2. Hyponasty
3. Epinasty
4. All of them

Answer: 4. All of them

Question 21. The closing of leaflets in Cisalpine is

1. Photonasty
2. Thermonasty
3. Nyctinasty
4. Seismonasty

Answer: 4. Seismonasty

Question 22. The primary root is:

Positively geotropic, Positively hydrotropic, Negatively geotropic, Negatively hydrotropic

1. 1, 3 Correct
2. 2, 4 Correct
3. 1, 2 Correct
4. 1, 2, 3 Correct

Answer: 3. 1, 2 Correct

Question 23. The opening and closing of a sunflower are also called

1. Photonasty
2. Chemonasty
3. Thermonasty
4. Seismonasty

Answer: 1. Photonasty

Question 24. The movement of tendrils of the pea is

1. Nutation
2. Circumnutation
3. Phototropism
4. Thigmotropism

Answer: 2. Circumnutation

Question 25. The movement that occurs in plants due to touch is

1. Thigmonasty
2. Thigmotropism
3. Both
4. None

Answer: 3. Both

Class 10 Life Science Chapter 1 Multiple Choice

Question 26. The response toward a stimulus is said to be

1. Negative
2. Avoidance
3. Positive
4. None of these

Answer: 3. Positive

WBBSE Chapter 1 Response And Chemical Coordination In Plant Hormones

Question 27. Who for the first time mentioned plant hormones?

1. Hugo de Vries
2. Charles Darwin
3. Gregor johann mendel
4. Robert Hooke

Answer: 2. Charles Darwin

Question 28. Which one is the natural auxin?

1. IAA
2. 2, 4-D
3. 2, 4, 5—T
4. Ga4

Answer: 1. IAA

Question 29. The hormone that helps in the ripening of fruit is

1. Florigen
2. Gibberellin
3. Ethylene
4. Auxin

Answer: 3. Ethylene

Class 10 Life Science Chapter 1 Important MCQs

Question 30. Which hormone is produced in the leaf but helps in flowering?

1. Phytochrome
2. Auxin
3. Gibberellin
4. Florigen

Answer: 3. Gibberellin

Question 31. Which hormone is produced in the coleoptile?

1. Auxin
2. Gibberellin
3. Cytokinin
4. Florigen

Answer: 1. Auxin

Question 32. Which hormone is used as a weedicide?

1. lAA
2. 2,4-D
3. GA
4. MCPA

Answer: 2. 2,4-D

Question 33. Name the hypothetical hormone that induces flowering.

1. Auxin
2. Kinetin
3. Ethylene
4. Florigen

Answer: 4. Florigen

Question 34. Name the hormone that helps in the growth of the axillary bud.

1. Auxin
2. Gibberellin
3. Kinetin
4. Florigen

Answer: 2. Gibberellin

Class 10 Life Science Chapter 1 Important MCQs

Question 35. The hormone that is produced in a maximum amount in the matured seed is

1. Auxin
2. Gibberellin
3. Kinetin
4. Zeatin

Answer: 2. Gibberellin

Class 10 Life Science Chapter 1 Questions For WBBSE

Question 36. The hormone that is present in coconut milk is

1. Kinetin
2. Auxin
3. Ethylene
4. None of them

Answer: 1. Kinetin

Question 37. The hormone that helps in the production of seedless fruit is

1. Auxin
2. Gibberellin
3. Kinetin
4. Ethylene

Answer: 3. Kinetin

Question 38. The following hormone does not contain nitrogen

1. 2,4-D
2. Iaa
3. Ga₃
4. Kinetin

Answer: 3. Ga₃

Class 10 Life Science Chapter 1 Important MCQs

Question 39. Substances that originate at the tip of the stem and control growth elsewhere are known as

1. Enzymes
2. Hormones
3. Vitamins
4. Growth inhibitors

Answer: 2. Hormones

Question 40. The following hormone is terpenoid in nature

1. Auxin
2. Gibberellin
3. Kinetin
4. Zeatin

Answer: 2. Gibberellin

Question 41. The hormone that is not a growth promoter

1. Auxin
2. Gibberellin
3. Abscisic acid
4. Kinetin

Answer: 3. Abscisic acid

Question 42. The hormone that helps in removing dwarfism is

1. Auxin
2. Gibberellin
3. Cytokinin
4. Abscisic acid

Answer: 2. Gibberellin

Question 43. Name the hormone whose precursor is tryptophan.

1. Auxin
2. Gibberellin
3. Cytokinin
4. Ethylene

Answer: 1. Auxin

Question 44. Name the hormone that induces apical dominance.

1. Auxin
2. Gibberellin
3. Cytokinin
4. Ethylene

Answer: 1. Auxin

Question 45. The hormone that may act as an anti-auxin

1. ABA
2. Ga₃
3. Florigen
4. Ethylene

Answer: 1. ABA

WBBSE Class 10 Life Science Chapter 1 Quiz

Question 46. Hormone primarily connected with cell division is

1. Ga
2. Cytokinin
3. Naa
4. IAA

Answer: 2. Cytokinin

Question 47. Name the non-indole auxin.

1. IAA
2. IBA
3. IPA
4. NAA

Answer: 4. NAA

Question 49. Which hormone is responsible for the Richmond lang effect?

1. Auxin
2. Gibberellin
3. Cytokinin
4. Ethylene

Answer: 3. Cytokinin

Question 50. Which one of the following is a rejuvenating hormone?

1. IAA
2. Ga₃
3. Kinetin
4. ABA

Answer: 3. Kinetin

Question 51. The hormone that is present in maize seed is

1. Kinetin
2. Zeatin
3. Iaa
4. Ga₃

Answer: 2. Zeatin

Question 52. The following hormone induces the closure of stomata

1. ABA
2. IBA
3. 2,4-D
4. Zeatin

Answer: 1. ABA

Question 53. Which is employed for the artificial ripening of banana fruit?

1. Auxin
2. Cytokinin
3. Coumarin
4. Ethylene

Answer: 4. Ethylene

WBBSE Class 10 Life Science Chapter 1 Quiz

Question 54. Apical dominance is caused by

1. Abscisic acid in lateral bud
2. Cytokinin in the leaf tip
3. Gibberellin in lateral bud
4. Auxin in shoot tip

Answer: 4. Auxin in shoot tip

Life Science MCQs For Class 10 WBBSE

Question 55. Flowering dependent on cold treatment is

1. Cryotherapy
2. Cryogenics
3. Cryoscopy
4. Vernalization

Answer: 4. Vernalization

Question 56. The translocation/movement of auxin is largely

1. Centripetal
2. Basipetal
3. Acropetal
4. Both 2 & 3

Answer: 4. Both 2 & 3

Question 57. Gibberellic acids induce flowering in

1. Short-day plants under long-day conditions
2. Long-day plants under short-day conditions
3. Day-neutral plants under dark conditions
4. Some gymnosperms only

Answer: 2. Long-day plants under short-day conditions

Question 58. Differentiation of shoot is controlled by

1. High cytokinin: auxin ratio
2. High auxin: cytokinin ratio
3. High gibberellin: auxin ratio
4. High cytokinin: gibberellin ratio

Answer: 1. High cytokinin: auxin ratio

WBBSE Class 10 Life Science Chapter 1 Quiz

Question 59. Pruning helps in making hedges dense because

1. It releases wound hormone
2. It induces the differentiation of new shoots from the rootstock
3. It frees axillary buds from apical dominance
4. The apical shoot grows faster after pruning

Answer: 3. It frees axillary buds from apical dominance

Question 60. Which one of the following pairs is not correctly matched?

1. Ga-leaf fall
2. Cytokinin-cell division
3. Laa-cell wall elongation
4. Aba-stomatal closure

Answer: 1. Ga-leaf fall

Question 61. Which of the following is a gaseous hormone?

1. IBA
2. NAA
3. Ethylene
4. ABA

Answer: 3. Ethylene

Question 62. Which of the following hormones is involved in counteracting apical dominance induced by auxin?

1. Cytokinin
2. Ethylene
3. Abscisic acid
4. Brassinosteroids

Answer: 1. Cytokinin

WBBSE Chapter 1 Response And Chemical Coordination In Animals Hormones MCQs

Question 63. Chemically hormones are

1. Lipids
2. Proteins
3. Carbohydrates
4. All of these

Answer: 4. All of these

Question 64. Response and coordination in animals involve the sense organs, nervous system, and chemicals called hormones.

1. Molecules
2. Substances
3. Messengers
4. Structures

Answer: 3. Messengers

Class 10 Life Science Chapter 1 MCQs With Answers

Question 65. Goosebumps are also called

1. Cutis anserine
2. Duckbumps
3. Coordination
4. Response

Answer: 1. Cutis anserina

Question 66. Thermo receptors are responsive to change of

1. Temperature
2. Humidity
3. Pressure
4. None of these

Answer: 1. Temperature

Question 67. An important hormone required for combatting stress is

1. Insulin
2. Thyroxin
3. Vasopressin
4. Adrenaline

Answer: 4. Adrenaline

Question 68. Hyperthyroidism results in

1. Virillism
2. Simple goitre
3. Exophthalmic goitre
4. Both (2) & (3)

Answer: 4. Both (2) & (3)

Question 69. The thyroid hormones t3 and i4 are the main regulators of

1. Stress
2. Blood pressure
3. Blood sugar
4. Bmr

Answer: 4. Bmr

Question 70. Estrogen and testosterone are

1. Growth hormones
2. Sex hormones
3. Emergency hormones
4. None of the above

Answer: 2. Sex hormones

Question 71. It stimulates the pituitary gland to release the sex hormones.

1. Testes
2. Ovary
3. Hypothalamus
4. Thyroid

Answer: 3. Hypothalamus

Question 72. Androgens are the hormones released by the

1. Testis
2. Pituitary
3. Adrenal
4. All of the above are correct

Answer: 1. Testis

Class 10 Life Science Chapter 1 MCQs With Answers

Question 73. Hormones are secreted by

1. Exocrine glands
2. Endocrine glands
3. Heterocrine glands
4. Both (2) & (3)

Answer: 4. Both (2) & (3)

Question 74. An example of an amine hormone is

1. Oestrogen
2. Oxytocin
3. Epinephrine
4. Follicle-stimulating hormone

Answer: 3. Epinephrine

Question 75. Examples of steroid hormones are

1. Thyroxine
2. Estrogens
3. Progesterone
4. Both ‘2’ and ‘3’ are correct

Answer: 4. Both ‘2’ and ‘3’ are correct

Question 76. The chemical structure of a hormone enables it to combine with a receptor in the cells of its

1. Target
2. Gland
3. Organ
4. All the above are correct

Answer: 1. Target

Question 77. Neurosecretory cells are found in

1. Liver
2. Hypothalamus
3. Kidney
4. Pancreas

Answer: 2. Hypothalamus

Question 78. Oxytocin and vasopressin are secreted by

1. Posterior pituitary
2. Anterior pituitary
3. Both ‘1’ and ‘2’
4. Neither

Answer: 1. Posterior pituitary

Class 10 Life Science Chapter 1 MCQs With Answers

Question 79. The hormones t3 and t4 work together to regulate the body’s

1. Temperature
2. Balance
3. Metabolic rate
4. Growth

Answer: 3. Metabolic rate

Question 80. The hormone epinephrine is produced from

1. Thymus
2. Adrenal medulla
3. Adrenal cortex
4. Thyroid

Answer: 3. Adrenal cortex

Question 81. The release of puberty triggers the development of female secondary sex characteristics.

1. Progesterone
2. Adrenaline
3. Fsh
4. Oestrogen

Answer: 4. Oestrogen

Question 82. During this time, the kidneys are unable to prevent the excretion of excess water.

1. Copd
2. Diabetes insipidus
3. Diabetes mellitus
4. Cretinism

Answer: 2. Diabetes insipidus

WBBSE Class 10 Life Science Chapter 1 MCQs

Question 83. Goitre is due to swelling of

1. Gall bladder
2. Pancreas
3. Testis
4. Thyroid

Answer: 4. Thyroid

Question 84. Is known as the master gland.

1. Pituitary
2. Thyroid
3. Ovary
4. Hypothalamus

Answer: 4. Hypothalamus

Question 85. Which one of the following is a local hormone?

1. Oxytocin
2. Insulin
3. Adrenaline
4. Gastrin

Answer: 4. Gastrin

Multiple Choice Questions On Control And Coordination

Question 86. Which hormone is not tropic?

1. ADH
2. STH
3. FSH
4. LH

Answer: 1. ADH

Question 87. Diabetes mellitus is associated with

1. Polyuria
2. Polydipsia
3. Polyphagia
4. All Of These

Answer: 4. All of the these

WBBSE Chapter 1 Response And Physical Coordination In Animals Nervous System MCQs

Question 88. In A Neuron, the Conversion Of Electrical Signal To Chemical Signal Occurs At/In

1. Axonal End
2. Dendritic End
3. Cell Body
4. Nodes Of Ranvler

Answer: 1. Axonal End

WBBSE Class 10 Life Science Chapter 1 MCQs

Question 89. The Nervous System Is Made Up Of Two Major Types Of Cells Known As

1. Axon And Dendron
2. Neuron And Nephron
3. Neurons And Neuroglia
4. Nucleus And Cytoplasm

Answer: 3. Neurons And Neuroglia

Question 90. Axons Are

1. Afferent
2. Efferent
3. Both ‘A’ And ‘B’
4. None

Answer: 2. Efferent

Question 91. Information From One Neuron Flows To Another Neuron Across A

1. Neuro-Muscular Junction
2. Neural Joint
3. Synapse
4. Ganglia

Answer: 3. Synapse

Question 92. Brain And Spinal Cord Constitute The

1. CNS
2. PNS
3. ANS
4. BMR

Answer: 1. CNS

Question 93. The Largest Part Of The Brain Is

1. Cerebellum
2. Cerebrum
3. Cerebral Cortex
4. Medulla

Answer: 2. Cerebrum

Class 10 Life Science Chapter 1 Multiple Choice

Question 94. Conditioned Reflexes Are

1. Inborn
2. Abnormal
3. Simple
4. Acquired

Answer: 4. Acquired

Question 95. The Thin, Transparent Tissue That Covers The Outer Surface Of The Eye Is

1. Conjunctiva
2. Cornea
3. Retina
4. Lens

Answer: 1. Conjunctiva

Question 96. The Accommodation Is A Of The Eye.

1. Part
2. Reflex Action
3. Both ‘A’ And ‘B’
4. None

Answer: 2. Reflex Action

Question 96. Near-sightedness problem Is Also Known As

1. Hypermetropia
2. Conjunctivitis
3. Myopia
4. Presbyopia

Answer: 3. Myopia

Question 97. A Bipolar Neuron Has

1. Only One Axon
2. One Axon & Many Dendrites
3. One Axon & One Dendron
4. Neither Any Dendron Nor Any Axon

Answer: 3. One Axon & One Dendron

Question 98. Motor Neurons Transmit Impulses

1. From Receptors To Cns
2. Spinal Cord To Brain
3. From Relay Neuron To Sensory Neuron
4. Cns To Effector

Answer: 4. Cns To Effector

Question 99. An Example Of Neurotransmitter Is

1. Lysozyme
2. Dopamine
3. Nissl Granules
4. Csf

Answer: 2. Dopamine

Class 10 Life Science Chapter 1 Multiple Choice

Question 100. The First And Second Lateral Ventricles Of The Brain Are Connected To The Third Ventricle Through The

1. Central Canal
2. Foramen Of Monro
3. Foramen Ovale
4. Foramen Magnum

Answer: 2. Foramen Of Monro

Question 101. Which Is The Correct Sequence Of The Components Of A Reflex Arc?

1. Receptors→Muscles→ Sensory Neuron→Motor Neuron→ Spinal Cord
2. Receptors→ Motor Neuron→ Spinal Cord→ Sensory Neuron→ Muscle
3. Receptors→ Spinal Cord→ Sensory Neuron→ Motor Neuron→ Muscle
4. Receptors→ Sensory Neuron→ Spinal Cord→ Motor Neuron→ Muscle

Answer: 4. Receptors→ Sensory Neuron→ Spinal Cord→ Motor Neuron→ Muscle

Objective Type Questions MCQs

Question 102. The part of the human brain associated with controlling body temperature is?

1. Thalamus
2. Cerebellum
3. Hypothalamus
4. Medulla oblongata

Answer: Hypothalamus.

Question 103. Which one of the following is the correct reflex arc?

1. Receptor → Effector → Efferent nerve → Nerve centre→ Afferent nerve
2. Nerve centre→ Receptor→ Afferent nerve→ Effector → Efferent nerve
3. Efferent nerve Receptor→ Afferent nerve→ Nerve centre→ Effector
4. Receptor → Afferent nerve → Nerve centre→ Efferent nerve→ Effector

Answer: Receptor → Afferent nerve → Nerve centre→ Efferent nerve → Effector

Class 10 Life Science Chapter 1 Important MCQs

Question 104. Determine the number of Cranial nerves in the human body —

1. 10 pairs
2. 31 pairs
3. 12 pairs
4. 21 pairs

Answer: 12 pairs.

WBBSE Chapter 1 Locomotion As A Type Of Response In Animals MCQs

Question 105. Amoeboid Locomotion Is Performed By

1. Cilia
2. Flagella
3. Muscles
4. Pseudopodia

Answer: 4. Pseudopodia

Question 106. Biped A I Ism Is Found In

1. Fishes
2. Birds
3. Euglena
4. Human

Answer: 4. Human

Question 107. Synovial Joints Are

1. Immovable
2. Movable
3. Slightly Movable
4. Both 2 And 3 Correct

Answer: 2. Movable

Question 108. Examples Of Hinge Joints Are

1. Shoulder
2. Skull
3. Knee
4. Vertebral Joints

Answer: 3. Knee

Question 109. Myotomes Are Found In

1. Fishes
2. Paramoecium
3. Euglena
4. All Animals

Answer: 1. Fishes

Class 10 Life Science Chapter 1 Important MCQs

Question 110. Pseudopodium In Ameoba Is Formed At The Anterior End By The

1. Conversion Of Plasmagel To Plasmasol
2. Flow Of Plasmasol At The Thinnest Point Of Cytoplasm
3. Conversion Of Plasmasol To Gel
4. None Of These

Answer: 3. Conversion Of Plasmasol To Gel

Question 111. Tail Feathers And Wing Feathers Of Birds Are Known As

1. Filoplumes And Contours
2. Rectrices And Remiges
3. Remiges And Rectrices
4. Contours And Rectrices

Answer: 2. Rectrices And Remiges

Question 112. Abductors Are Antagonists Of

1. Adductors
2. Extensors
3. Flexors
4. None Of These

Answer: 1. Adductors

Question 113. When A Joint Is Such That It Allows the Movement Of Bones In All Directions, It Is Known As A

1. Hinge Joint
2. Angular Joint
3. Pivot Joint
4. Ball And Socket Joint

Answer: 4. Ball And Socket Joint

Question 114. Elbow Joint Is

1. Pivot Joint
2. Hinge Joint
3. Gliding Joint
4. None Of These

Answer: 2. Hinge Joint

Question 115. Synovial Fluid Is Secreted By

1. Cartilage
2. Bones
3. Synovial Membrane
4. All Of These

Answer: 3. Synovial Membrane

Class 10 Life Science Chapter 1 MCQs With Answers

Question 116. Which one is the largest synovial joint of the human body?

1. Shoulder joint
2. Knee joint
3. Hip joint
4. Elbow joint

Answer: 3. Hip joint

Question 117. A sessile animal Is

1. Starfish
2. Earthworm
3. Coral
4. Snail

Answer: 3. Coral

Question 118. Flight muscles of birds are attached to

1. Keel of sternum
2. Clavicle
3. Pelvic girdle
4. Humerus

Answer: 1. Keel of sternum

Question 119. Some flowers are open after sunrise and close after sunset. This is

1. Photonasty
2. Seismonasty
3. Chemonasty
4. Thermonasty

Answer: 1. Photonasty

Question 120. The tentacles on the leaves of an insectivorous plant called sundew bend to trap the insect as soon as they come in contact with the insect’s body. This is

1. Seismonasty
2. Thermonasty
3. Photonasty
4. Chemonasty

Answer: 4. Chemonasty

Question 121. Select which of the following statements is correctly related to tropic movement-

1. It is controlled by the intensity of the stimulus
2. The overall change of place takes place by a plant or part of a plant
3. This movement is observed in the algae called volvox
4. It is an induced movement of curvature controlled by the direction of the stimulus

Answer: 1. It is an induced movement of curvature controlled by the direction of the stimulus

Model Question Paper Life Science Set 2 Group A

Answer to all questions is compulsory

Write the answer in a complete sentence by choosing the correct answer for each question with the respective serial number.

Question 1. Select the correct pair-

1. Sclera—Absorbs excess light within the eyeball
2. Choroid—Provides definite shape of the eyeball
3. Lens-Causes refraction of light and accomplishes accommodation
4. Retina-Holds the lens with the help of the suspensory ligaments

Answer: 3. Lens-Causes refraction of light and accomplishes accommodation

Question 2. Identify which of the following statements is not true regarding Adrenaline-

1. Increases the rate of heartbeat
2. Helps in the production of spermatozoa during adolescence
3. Increases cardiac output
4. Increases systolic blood pressure

Answer: 2. Helps in the production of spermatozoa during adolescence

Question 3. Select which of the following is the feature of tropic movement-

1. Induced movement of curvature of particular parts of the plant controlled by the direction of the stimulus
2. It is one type of turgor movement
3. Induced movement of curvature of particular parts of plant controlled by the intensity of the stimulus
4. Does not occur under the influence of Auxin

Answer: 1. Induced movement of curvature of particular parts of the plant controlled by the direction of the stimulus

Question 4. Match the words of Column-A with that of Column-B and select which of the following options is correct—
Answer: table-

1. A-2,B-3,C-1
2. A-3,B-1,C-2
3. A-1,B-2,C-3
4. A-1,B-3,C-2

Answer: 2. A-1,B-3,C-2

WBBSE Class 10 Life Science Model Question Paper 2023

Question 5. Decide which of the following sequence is correct-

1. Receptor→Effector→Sensory Nerve→Motor Nerve→Nerve Centre
2. Receptor→Nerve Centre→Motor Nerve→Sensory Nerve→Effector
3. Receptor→Sensory Nerve→Nerve Centre→Motor Nerve→Effector
4. Receptor→Motor Nerve→Effector→Sensory Nerve→Nerve Centre

Answer: 3. Receptor→Sensory Nerve→Nerve Centre→Motor Nerve→Effector

Question 6. Select the correct pair-

1. Telophase- Movement of daughter chromosomes towards poles
2. Telophase- Disappearance of Nuclear Membrane and Nucleolus
3. Telophase- Formation of spindle fibre
4. Telophase- Reappearance of Nuclear Membrane and Nucleolus

Answer: 4. Telophase- Reappearance of Nuclear Membrane and Nucleolus

Question 7. The name of the terminal parts of the Chromosome is—

1. Centromere
2. Telomere
3. Nucleolar Organizer
4. Satellite

Answer: Telomere

Question 8. Identify which of the following statements is not true regarding cross-pollination-

1. Requires agent
2. Origin of new hereditary features
3. Maintains the purity of species
4. The rate of germination of seeds is higher

Answer: 3. Maintains the purity of species

Question 9. Consider the differences between asexual and sexual reproduction and select which of the options is/are correct-
Answer:

1. 1,4
2. 2,3
3. 3,4
4. 1,2

Answer: Correct option is not provided

Question 10. Select the correct pair in the case of humans:

1. The normal pattern of the chromosome of ovum — 22A + XX
2. The normal pattern of the chromosome of the ovum — 22A + Y
3. The normal pattern of the chromosome of ovum — 22A + X
4. The normal pattern of the chromosome of ovum — 22A + XY

Answer: 3. Normal pattern of the chromosome of ovum — 22A + X

WBBSE Class 10 Life Science Model Question Paper 2023

Question 11. 1.11Determine which of the following genotypes is homozygous for both the loci

1. BbRr
2. BBRr
3. BbRR
4. bbrr

Answer: 4. bbrr

Question 12. Identify the genotype of guineapig having white coat colour and rough hair-

1. bbRR,brr
2. BBRR,brr
3. bb,bb
4. BbRr, BbRR

Answer: 3. bbRR, bbRr

Question 13. Decide which of the following two were selected by Mendel as dominant traits—

1. Length of stem — Long, Shape of seed — Round
2. Position of flower — Terminal, the colour of flower — White
3. Colour of the cotyledon — Green, and Shape of the seed – Wrinkled
4. Length of stem — Dwarf, Shape of seed – Wrinkled

Answer: 1. Length of stem — Long, Shape of seed — Round

Question 14. Determine the nature of the pea plant germinated from the seed produced as a result of pollination between a hybrid tall (Tt) and a pure dwarf (tt) pea plant—

1. 100% Tall
2. 50% Tall, 50% Dwarf
3. 100% Dwarf
4. 75% Tall, 25% Dwarf

Answer: 2. 50% Tall, 50% Dwarf

Question 15. Decide which of the following diseases can be prevented by taking suggestions from pre-marital genetic counselling-

1. Goitre
2. Malaria
3. Thalassaemia
4. Tuberculosis

Answer: 3. Thalassaemia

Model Question Paper Life Science Set 2 Group B

2. Answer 21 questions out of 26 questions given below as instructed. Fill in the blanks with proper words in the following sentences (any five):

Question 1. The hormone is found in very high amounts in the ripe seeds of plants.
Answer: Gibberellin

Question 2. Reflex action is fast, spontaneous and 
Answer: Involuntary

Question 3. In cell division spindle fibres are not formed.
Answer: amitosis

Question 4. is an entomophilous flower.
Answer: Sunflower

WBBSE Class 10 Life Science Model Question Paper 2023

Question 5. The gene having the power of the rolling tongue is.
Answer: dominant

Question 6. As the gene responsible for hemophilia is recessive, the symptoms of this disease are only manifested in
condition.
Answer: X-linked inheritance

Decide whether the following statements are true or false (any five):

Question 7. The focal length of the lens increases in case of seeing distant objects.
Answer: True

Question 8. Each nucleoside contains a nitrogenous base and phosphoric acid.
Answer: False

Question 9. There is no role of females in determining the sex of humans.
Answer: True

Question 10. If any pea plant contains either TT or tt trait, then the pea plant becomes heterozygous with respect to such alleles.
Answer: False

Question 11. As the flowers of the pea plant are unisexual, so self-pollination and if needed, cross-pollination can be performed.
Answer: True

Question 12. The daughter cells are changed and modified in different ways during cell differentiation phase to form tissue, organ and system.
Answer: True

WBBSE Class 10 Life Science Model Question Paper 2023

Match the words in Column-A with those which are most appropriate in Column-8 and re-write the correct pair mentioning the serial no. of both columns (any five) :

Answer: 2.13 – (2), 2.14- (3) 2.15- (4), 2.16- (5), 2.17- (6), 2.18- (1)

Answer in a single word or in a single sentence (any six):

Question 19. Choose the odd one and write it: Olfactory nerve, Vagus nerve, Optic nerve, Auditory nerve
Answer: Vagus nerve

Question 20. What is the fate of hormones in the animal body after its function is over ?
Answer: Hormones are destroyed and excreted immediately after their functions are over.

Question 21. A pair of related terms is given below. On the basis of the relationship in the first pair write the suitable word in the I gap of the second pair. Purine : Adenine :: Pyrimidine :
Answer: Cytosine

Question 22. What happens if the functions of checkpoints in the cell cycle get hampered?
Answer: Genomic instability and/or tumorigenesis

Question 23. What is hybridization?
Answer: Process of interbreeding individuals from genetically distinct populations to produce hybrids.

Question 24. What is a dominant trait?
Answer: A trait produced by a dominant allele in individuals who have one copy of the allele which can come from just one parent.

Class 10 Life Science Model Question Paper With Answers

Question 25. Among the following four terms, one includes the other three. Find it out and write it :
Increase in Basal Metabolic Rate, Thyroxine, Gradual maturation of Red Blood Corpuscle, Exophthalmic Goitre.
Answer: Exophthalmic Goitre

Question 26. In which phase of Interphase are the proteins essential for the formation of spindle fibres in synthesized?
Answer: G2 – phase.

Extensor muscle:

to increase the angle between two body parts.

Abductor’s muscle:

to move limbs away from the midline.

Rotator muscle:

To move limbs around their long axis.

Question 27. Prepare a list of functions controlled by Auxin related to the growth of plants.
Answer:

Major Plant Hormones Auxin

Auxins are a class of nitrogenous phytohormones & the most important growth regulators produced from the growing regions of plant tissue including roots, shoots, and leaves.

These promote cell division, stem and root growth. These can also drastically affect plant orientation by promoting cell division on one side of the plant in response to sunlight & gravity.

The term auxin was first used by Frits Went.

Chemical composition:

Auxins generally occur as complexes, usually found with an amino acid or sugar. These are composed of carbon, hydrogen, oxygen, and nitrogen.

The amino acid Tryptophan and many other Indole compounds serve as precursors of auxins.

Class 10 Life Science Model Question Paper With Answers

Chemical composition:

Auxins generally occur as complexes, usually found with an amino acid or sugar. These are composed of carbon, hydrogen, oxygen, and nitrogen.

The amino acid Tryptophan and many other Indole compounds serve as precursors of auxins.

Types: They are classified into two types as detailed below:

Natural Auxin:

They are Indole compounds like Indole 3-Acetic Acid (IAA) and Non-Indole compounds like Naphthalene Acetic Acid. Indole 3-Acetic Acid was the first plant hormone identified.

It is manufactured primarily in the shoot tips (in leaf primordia and young leaves), in embryos, and in parts of developing flowers and seeds.

Synthetic Auxin:

These are synthetic compounds similar to natural auxin and they include 2,4-D (2,4-Dichlorophenoxy Acetic Acid) or 2, 4, 5-T (2,4,5- Trichlorophenoxy Acetic Acid).

Translocation:

The transport of IAA from cell to cell through the parenchyma surrounding the vascular tissues requires the expenditure of ATP energy.

IAA moves in one direction only i.e., the movement is polar and in this case, downward.

Such downward movement in shoots is said to be basipetal (apex to base) and in roots it is acropetal (outwards towards the root apices from the base).

Synthetic auxins move in all directions inside plants.

Function:

Apical Growth:

Auxins bring about apical dominance and thereby induce apical growth of the shoot.

Tropic movements:

These are deposited unequally on the shoot or root surface and thus cause phototropic curvature and geotropism

Cell enlargement:

Auxin helps to increase the size and volume of the cells.

Cell enlargement is caused by the solubilization of carbohydrates, loosening of wall microfibrils, synthesis of more wall materials, increased membrane permeability, and respiration.

Metabolism:

Auxin stimulates respiration by increasing the availability of respiratory substrate. Thus it enhances metabolism by mobilizing plant resources.

Cambial activity:

The degree of cambial activity is directly proportional to auxin concentration. Auxin also controls xylem differentiation.

Cell division:

It initiates the cell division of the parenchymatous cells at the site of wounded tissue causing healing of wounds after mechanical injury in plants.

Class 10 Life Science Model Paper WBBSE 2023 Set 2

Root formation:

Auxin promotes root initiation at a concentration that is inhibitory for the growth of intact roots.

Parthenocarpic fruit Development:

Pollen contains a lot of auxin, it acts as a signal indicating the completion of pollination and transformation of the ovary to fruit.

Thus the application of synthetic auxin induces the development of fruit without the act of fertilization i.e. parthenocarpic fruit development.

Synthetic auxin (2,4-D) initiates rootings in stem cuttings.

(x)Synthetic auxin can act as a weedicide by killing dicotyledonous weeds preventing the division of cambial tissue.

As monocotyledonous plants are devoid of cambium, they are not destroyed by auxin compounds.

Anti-auxin:

Auxins have an inhibitory effect on the abscission of leaves and fruits, i.e., shedding of mature leaves and ripe fruits from the plant.

It has been found that the abscission zone does not form when the concentration of auxins is high in the leaves or fruits.

On the contrary, Abscisic acid (ABA) is a major hormone in plants that induces the formation of an abscission zone in the leaf stalk or petiole and brings about the shedding of leaves.

It acts as an antiauxin because it counteracts the auxin activity by initiating abscission.

In Plants Hormones Gibberellin

Gibberellins are one of the longest-known classes of non-nitrogenous phytohormones that regulate various developmental processes including stem elongation, germination, dormancy, flowering, and leaf and fruit senescence.

Gibberellic acid or gas can be of more than 100 types and the most common being 3- Gibberellic Acid (GA₃).

Gibberellin was discovered by Kurosawa (1926) from rice plants suffering from Bakanae (foolish seedling) disease caused by Gibberella.

Gibberellins are naturally synthesized in maturing seeds, germinating seedlings, growing tissues of expanding cotyledons, growing leaves, apical buds, root tips, etc.

Class 10 Life Science Model Paper WBBSE 2023 Set 2

Chemical composition:

Gibberellins are tetracyclic diterpene acids. These consist of carbon, hydrogen, and oxygen.
Acetate is the precursor for the synthesis of all kinds of gibberellins.

The gibberellins are named GAj through GAn in the order of discovery.

Translocation:

All known gibberellins are synthesized in plastids and then modified in endoplasmic reticulum and cytosol until they reach their biologically active form.

The transport of gibberellin is not polar. These are transported in all directions from the site of formation through the xylem, phloem, or by cell tq cell.

Functions:

Elongation of internode:

It brings about the growth of the internode and thereby brings about apical growth along with auxin.

Flowering:

It modifies the apical bud to the floral bud, by bringing about the development of floral tissue through modification of the apical meristem.

Breaks seed dormancy:

It breaks seed dormancy and induces germination of seed through denovo synthesis of a-amylase.

Leaf growth:

It helps in the development of young leaves through stimulation of the leaf meristem

It promotes fruit growth and development. It also influences parthenocarpy.

Question 28. Distinguish between the mitosis of plant cells with the mitosis of animal cells on the basis of the following features—

1. Formation of spindle fibers
2. Process of cytokinesis

Answer:

Basic Differences Between Mitosis Of Animal Cell And Plant Cell

Model Question Paper Life Science Set 2 Group C

3. Answer any 12 questions in 2-3 sentences out of 17 questions given below:

Question 1. Prepare a list of hormones that control the following functions—

Causes an increase in the length of bones through mineralization of the matrix of cartilage located at the terminal parts of long bones.

Answer: GH or STH

Transforms ruptured follicle into a temporary endocrine gland namely Corpus Luteum and provides necessary stimulation for the secretion of Progesterone hormone from that gland.

Answer: LH

Inhibits the production of glucose in the liver from protein and lipids.
Answer: Insulin

Causes contraction of uterine muscles during parturition
Answer: Relaxin

Class 10 Life Science Model Paper WBBSE 2023 Set 2

Question 2. Explain the causes of Myopia and Hyperopia
Answer:

Endocrine disorders or hormonal disorders are typically an endocrine disease that results when a gland produces too much (hyper-secretion) or too little (hypo-secretion) of an endocrine hormone, called a hormone imbalance.

Some important hormonal disorders of humans are—

Dwarfism:

Pituitary dwarfism is decreased bodily growth due to hormonal problems (hyposecretion of STH) in childhood.

Causes:

Pituitary dwarfism, or growth hormone deficiency, is a condition in which the pituitary gland does not make enough growth hormone (hypo-function of STH).

Pituitary gland dysfunction can be congenital, which means that the child is born with the abnormality, or can be acquired during or after birth. It tends to run in families.

Symptoms:

The main symptom of pituitary dwarfism is below-average growth, although body proportions will be normal. Other characteristics might include an immature appearance, a chubby body build, a prominent forehead, and an underdeveloped bridge of the nose.

Diabetes insipidus (Dl) is a condition in which the kidneys are unable to prevent the excretion of water due to the hyposecretion of ADH leading to the excretion of an abnormally large amount of urine from the body.

Causes:

During the day, our kidneys filter the blood many times. Normally, most of them are reabsorbed and only a small amount of concentrated urine is excreted.

Dl occurs when the kidneys cannot concentrate the urine normally due to a decrease in the reabsorption of water in the renal tubules and a large amount of dilute urine is excreted.

The amount of water excreted in the urine is controlled by antidiuretic hormone (ADH). Dl caused by a lack of ADH is called central diabetes insipidus.

Excessive thirst that may be intense or uncontrollable, usually with the need to drink large amounts of water.

Excessive urine volume,

Excessive urination, often needing to urinate every hour throughout the day and night.

Goitre:

Goitre is a swelling (hypertrophy) of the thyroid gland in the neck due to hypothyroidism.

Causes:

Iodine deficiency, leading to hypothyroidism, is the major cause of endemic goiter. The thyroid gland needs iodine in order to manufacture thyroid hormones, which regulate the body’s rate of metabolism.

Hypothyroidism is the result of an underactive thyroid gland, and this causes goiter. Because the gland produces too little thyroid hormone, it is stimulated to produce more, leading to swelling.

Symptoms:

The main symptom of goiter is swelling of the thyroid gland, which causes a lump to develop in the front of the neck.

The following are the main symptoms that can result from neck swelling- symptoms of tightness, cough, and hoarseness; Trouble swallowing (dysphagia), dry & rough skin, hair loss, excessive fatigue, etc.

Class 10 Life Science Model Paper WBBSE 2023 Set 2

Diabetes mellitus:

Diabetes mellitus is a chronic, lifelong condition that affects the body’s ability to use the energy found in food.

There are two major types of diabetes:

1. Type-1 diabetes and
2. Type-2 diabetes.

Type-1 diabetes is also called insulin-dependent diabetes. It used to be called juvenile-onset diabetes because it often begins in childhood.

Type-2 diabetes used to be called adult-onset diabetes, but with the epidemic of obese and overweight kids, more teenagers are now developing type-2 diabetes.

Type-2 diabetes is also called non-insulin-dependent diabetes.

Causes:

All types of diabetes mellitus have something in common. Normally, our body breaks down the sugars and carbohydrates we eat into glucose.

Glucose fuels the cells in our body. But the cells need insulin, a hormone, in the bloodstream in order to take the glucose and use it for energy.

With diabetes mellitus, either the body doesn’t make enough insulin or it can’t use the insulin it produces or a combination of both.

In Type-2 diabetes, the pancreas usually produces some insulin. But either the amount produced is not enough for the body’s needs, or the body’s cells are resistant to it.

Symptoms:

Since the cells cannot take in the glucose, it builds up in the blood. High levels of FuncIaivientaIs of UFe Sconce
blood glucose can damage the tiny blood vessels in the kidneys, heart, eyes, or nervous system.

That is why diabetes – especially if left untreated – can eventually cause heart disease, stroke, kidney disease, blindness, and damage to nerves in the feet.

The early symptoms of untreated diabetes are related to elevated blood sugar levels and loss of glucose in the urine. The high amount of glucose in the urine can cause increased urine output and leads to dehydration.

This condition is called Polyuria. Dehydration causes increased thirst and water consumption. This condition is known as polydipsia.

The inability of insulin to perform normally has effects on protein, fat, and carbohydrate metabolism. A relative or absolute insulin deficiency eventually leads to weight loss despite an increase in appetite.

This condition is called polyphagia. Some untreated diabetes patients also complain of fatigue, nausea, and vomiting. Fluctuations in blood glucose levels can lead to blurred vision.

Extremely elevated glucose levels can lead to lethargy and coma.

The Common Hormonal Disorders Are Summarised In The Following Table:

Question 3. Explain what phenomena happen when the following muscles contract :

1. Flexor muscle :
2. Extensor muscle :
3. Abductor’s muscle :
4. Rotator muscle :

Answer:

Flexor muscle :

To decrease the angle between two body parts.

Extensor muscle :

To increase the angle between two body parts.

Abductor’s muscle :

To move limbs away from the midline.

Rotator muscle :

To move limbs around their long axis.

WBBSE Life Science Class 10 Question Paper Set 2

Question 4. Prepare a list of functions controlled by Auxin related to the growth of plants.
Answer:

Major Plant Hormones Auxin

Auxins are a class of nitrogenous phytohormones & the most important growth regulators produced from the growing regions of plant tissue including roots, shoots, and leaves.

These promote cell division, stem and root growth. These can also drastically affect plant orientation by promoting cell division on one side of the plant in response to sunlight & gravity.

The term auxin was first used by Frits Went.

Chemical composition:

Auxins generally occur as complexes, usually found with an amino acid or sugar. These are composed of carbon, hydrogen, oxygen, and nitrogen.

The amino acid Tryptophan and many other Indole compounds serve as precursors of auxins.

Chemical composition:

Auxins generally occur as complexes, usually found with an amino acid or sugar. These are composed of carbon, hydrogen, oxygen, and nitrogen.

The amino acid Tryptophan and many other Indole compounds serve as precursors of auxins.

Types: They are classified into two types as detailed below:

Natural Auxin:

They are Indole compounds like Indole 3-Acetic Acid (IAA) and Non-Indole compounds like Naphthalene Acetic Acid. Indole 3-Acetic Acid was the first plant hormone identified.

It is manufactured primarily in the shoot tips (in leaf primordia and young leaves), in embryos, and in parts of developing flowers and seeds.

Synthetic Auxin:

These are synthetic compounds similar to natural auxin and they include 2,4-D (2,4-Dichlorophenoxy Acetic Acid) or 2, 4, 5-T (2,4,5- Trichlorophenoxy Acetic Acid).

Translocation:

The transport of IAA from cell to cell through the parenchyma surrounding the vascular tissues requires the expenditure of ATP energy.

IAA moves in one direction only i.e., the movement is polar and in this case, downward.

Such downward movement in shoots is said to be basipetal (apex to base) and in roots it is acropetal (outwards towards the root apices from the base).

Synthetic auxins move in all directions inside plants.

Function:

Apical Growth:

Auxins bring about apical dominance and thereby induce apical growth of the shoot.

Tropic movements:

These are deposited unequally on the shoot or root surface and thus cause phototropic curvature and geotropism

Cell enlargement:

Auxin helps to increase the size and volume of the cells.

Cell enlargement is caused by the solubilization of carbohydrates, loosening of wall microfibrils, synthesis of more wall materials, increased membrane permeability, and respiration.

Metabolism:

Auxin stimulates respiration by increasing the availability of respiratory substrate. Thus it enhances metabolism by mobilizing plant resources.

Cambial activity:

The degree of cambial activity is directly proportional to auxin concentration. Auxin also controls xylem differentiation.

Cell division:

It initiates the cell division of the parenchymatous cells at the site of wounded tissue causing healing of wounds after mechanical injury in plants.

Root formation:

Auxin promotes root initiation at a concentration that is inhibitory for the growth of intact roots.

Parthenocarpic fruit Development:

Pollen contains a lot of auxin, it acts as a signal indicating the completion of pollination and transformation of the ovary to fruit.

Thus the application of synthetic auxin induces the development of fruit without the act of fertilization i.e. parthenocarpic fruit development.

Synthetic auxin (2,4-D) initiates rootings in stem cuttings.

(x)Synthetic auxin can act as a weedicide by killing dicotyledonous weeds preventing the division of cambial tissue.

As monocotyledonous plants are devoid of cambium, they are not destroyed by auxin compounds.

Anti-auxin:

Auxins have an inhibitory effect on the abscission of leaves and fruits, i.e., shedding of mature leaves and ripe fruits from the plant.

It has been found that the abscission zone does not form when the concentration of auxins is high in the leaves or fruits.

On the contrary, Abscisic acid (ABA) is a major hormone in plants that induces the formation of an abscission zone in the leaf stalk or petiole and brings about the shedding of leaves.

It acts as an antiauxin because it counteracts the auxin activity by initiating abscission.

WBBSE Life Science Class 10 Question Paper Set 2

Question 5. Distinguish between the mitosis of plant cells with the mitosis of the animal cell on the basis of the following features—

1. Formation of spindle fibers
2. Process of cytokinesis

Answer:

Basic Differences Between Mitosis Of Animal Cell And Plant Cell

Question 6. Explain how Bryophyllum undergoes vegetative propagation with the help of an adventitious bud.
Answer:

Vegetative Reproduction

Vegetative reproduction is basically a special type of asexual reproduction where a vegetative part of the plant body, separated from the original plant body, develops and grows into a new plant by simple cell division.

The Vegetative Reproduction Is Of Two Kinds:

1. Methods of natural vegetative reproduction and
2. Methods of artificial vegetative reproduction.

It occurs naturally through budding, fission, fragmentation, etc. Artificially this reproduction can be carried out by cutting, grafting, etc.

Parthenogenesis or parthenocarpy

Parthenogenesis is a form of asexual reproduction in which the growth and development of the embryo occur without fertilization.

In animals, parthenogenesis means the development of an embryo from an unfertilized egg cell. It is common in rotifers, aphids, bees, and crustaceans.

Some vertebrates like lizards also reproduce by parthenogenesis. Parthenogenesis may be natural or artificial.

Natural parthenogenesis occurs regularly in the life cycle of some animals. It is of two types:

Complete and incomplete parthenogenesis.

Complete parthenogenesis:

It is also called obligatory partheno-genesis.

In this case, the males are completely absent and the females develop from the unfertilized eggs as is found in aphids, phyllopods, and rotifers.

It is also found in some vertebrates.

A lizard Lacerta Mexico Americana reproduces exclusively by parthenogenesis with no males, in the population.

Incomplete parthenogenesis:

In this case sexual generation alternates with parthenogenesis generation.

For example, in bees and wasps, some eggs develop without fertilization and produce males, while those eggs that are fertilized develop into females.

In Gall fly, the larvae may lay eggs which develop parthenogenetically into a new generation of larvae. This is called paedogenetic parthenogenesis or paedogenesis.

In many sexually reproducing animals, the egg can be activated by artificial methods to start the development without fertilization.

This is called artificial parthenogenesis. Eggs of Sea urchins can be made to develop successfully if treated with weak salt solutions, weak organic acids, electric shock or shaking in seawater, or by pricking the egg with a glass needle.

In higher plants, the process of formation of fruits without pollination and fertilization is called parthenocarpy. Parthenocarpy is a form of asexual reproduction seen in flowering plants.

The fruits are generally seedless. During cultivation, parthenocarpy is introduced along with other plant hormones including gibberellin and it results in maturing of the ovaries without the process of fertilization and produces bigger and pulpy fruits.

Pineapples, bananas, cucumber, grapes, watermelon, oranges, pears,s, etc are some examples of parthenocarpy. In some plants, pollination or another stimulation is required for parthenocarpy. This is termed stimulative parthenocarpy.

Plants that do not require pollination or other stimulation to produce parthenocarpic fruits have vegetative parthenocarpy.

Seedless cucumbers are an example of vegetative parthenocarpy and seedful watermelon is an example of stimulative parthenocarpy.

WBBSE Life Science Class 10 Question Paper Set 2

Question 7. Tabulate which changes occur in a chromosome during the Anaphase of Karyokinesis.
Answer:

Changes occur in a chromosome during the Anaphase of Karyokinesis

1. Splitting into daughter chromosomes.
2. Poleward anaphase movement of daughter chromosomes.
3. V, L, or I shaped the appearance of daughter chromosomes.

Question 8. Describe how Yeast completes the process of budding.
Answer:

Yeast completes the process of budding

The asexual reproduction of yeast occurs by budding.

The parent nucleus divides into two daughter nuclei by mitosis.

During budding, the daughter cell first appears as an outgrowth since one daughter nucleus migrates to a corner of the parent cell. Both haploid and diploid cells can undergo budding. The outgrowth enlarges, matures, and detaches from the parent cell.

Question 9. Prepare a list of changes that occur during the adolescence phase of human development
Answer:

Changes that occur during the adolescence phase of human development

1. The rate of growth becomes rapid.
2. Secondary sexual characters develop.
3. Significant growth of sexual organs takes place.
4. The thought process becomes logical and complex.

Question 10. Explain the following two significances of meiotic cell division—
Maintains the constant number of chromosomes in a species Origin of variation in an organism
Answer: 

Significance Of Meiosis

1. Gametogenesis:

It is a necessary part of the life cycle of sexually reproducing animals since it leads to the formation of gametes (sex cells) through meiosis.

2. Maintenance Of Constant Chromosome Number Of A Species:

The gametes produced as a result of meiosis are haploid (n) and the zygote formed by their fusion is diploid (2n).

Thus, it is the only means for restoring the constancy in chromosome number, which is a characteristic of the species from generation to generation.

3. Production Of Variation In Organisms:

Meiosis results in new combinations of genetic material. During crossing over, the hereditary factors from male and female parents get mixed due to the breakage and exchange of chromatids.

This brings in genetic variation within the species. The variations are important raw materials for evolution and also help in the improvement of races.

4. Alternation of generation:

Meiosis causes conversion from sporophytic generation to gametophytic generation in plants, i.e. causes the alternation of generation through prezygotic, post-zygotic, or sporadic meiosis.

Meiosis occurs in sporogenous cells (micro and megaspore mother cells) of the sporophyte-producing haploid spores. The spores on germination form gametophytes (male and female).

Cells of the gametophyte form gametes. Fusion of these gametes again leads to diploid or sporophytic generation and in this way, the alternation of generation between gametophytic and sporophytic generations keeps on going.

Question 11. State Mendel’s second law of heredity.
Answer:

Mendel’s Fourth Postulate

The above analysis became the basis of Mendel’s second general principle or the fourth postulate, the law of independent assortment.

Independent Assortment:

During gamete formation, segregating pairs of unit factors assort independently of each other and undergo random recombination in all possible combinations governed by chance alone.

This postulate stipulates that any pair of unit factors segregate independently of all other unit factors. Thus, according to the postulate of independent assortment, all possible combinations of gametes will be formed in equal frequency.

Question 12. Prove the authenticity of the statement- “The phenotypic and genotypic ratio remain identical in case of Incomplete Dominance”.
Answer:

Incomplete Dominance

A common example of deviation from Mendelism is the phenomenon called incomplete dominance.

A cross between parents with contrasting traits may sometimes generate offspring with an intermediate phenotype.

In a heterozygote organism carrying both a dominant and a recessive allele of the same gene, when the dominant gene cannot express its dominant phenotype completely, a mixed or intermediate, or blended phenotype between the dominant and the recessive is expressed.

Such a situation is known as incomplete dominance. In many plant species, flower color serves as a striking example of incomplete dominance.

With the flowers of Four O’Clocks or floret clusters of Snapdragons Mirabilis jalapeno, a cross between pure breeding red flowered parents and pure breeding white yields hybrids with pink blossoms.

During gametogenesis, the pure red flowered parent plant (AA) produces (A) gametes and the pure white flowered parent plant (aa) produces (a) gametes.

After cross-pollination of parental plants, (A) and (a) unite together to form (Aa) zygote that develops into plants with pink flowers.

Here both the allelomorphic genes have a partial or incomplete dominant relationship and hence, F₁ hybrids show a mixture of characters of both parents. This is a case of incomplete dominance.

If allowed to self-pollinate, the F₁ pink blooming plants produce F₂ progeny bearing red, pink, and white flowers in a ratio of 1: 2 :1. This is the familiar genotypic ratio of an ordinary single gene F₁ self-cross.

What is new is that because the heterozygotes look unlike either homozygote, the phenotypic ratios are an exact reflection of the genotypic ratios.

F₂ Phenotypic ratio = 1 (Red): 2 (Pink): 1 (White) F₂ Genotypic ratio = 1 (AA): 2 (Aa): 1 (aa) In this example of Mirabilis jalapa, the red gene is incompletely dominant over the white gene and so both of them give rise to an intermediate pink colored flower in heterozygous or hybrid condition.

WBBSE Life Science Class 10 Question Paper Set 2

Explanation:

The biochemical explanation for this type of incomplete dominance is that each allele of the gene under analysis specifies an alternative form of a protein molecule with an enzymatic role in pigment production.

If the ‘white’ allele does not give rise to the functional enzyme, no pigment appears.

Thus, in Snapdragons and four o’clock, two ‘red’ alleles per cell produce a double dose of a red-producing enzyme, which generates enough pigment to make the flowers look fully red.

In the heterozygote, one copy of the ‘red’ allele per cell results in only enough pigment to make the flowers look pink. In the homozygote for the ‘white’ allele, where there is no functional enzyme and thus no red pigment, the flowers appear white.

Question 13. Write two causes behind Mendel’s success in conducting experiments on heredity.
Answer:

Reasons for Mendel’s Success

Mendel’s success was dependent upon the following factors:

First, he chose the garden pea (Pisum sativum) as his experimental organism.

These plants can easily be cultivated, crossed, and for each successive generation, Mendel could thus obtain large members of individuals within a relatively short growing season.

By comparison, if he had worked with sheep, each mating would have generated only a few offspring and the time between generations would have been several years.

Second, Mendel examined the inheritance of clearcut contrasting forms of particular traits — round versus wrinkled seed, yellow versus green pod color, etc.

Using such ‘either-or’ traits, he could distinguish and trace unambiguously the transmission of one or the other observed characteristics, because there were neither any intermediate forms nor any of these characters located on separate chromosomes.

Third, Mendel isolated and perpetuated lines of peas that breed true. Mating with such pure breeding lines produce offsprings carrying specific parental traits that remain constant from generation to generation.

Fourth, Mendel carefully controlled his matings, going to great lengths to ensure that the progeny he observed really resulted from the specific fertilization he intended.

Thus he painstakingly prevented the intrusion of any foreign pollen and assured self or cross-pollination as the experiment demanded.

He also performed reciprocal crosses, in which by reversing the traits of male and female parents, he efficiently controlled the path of transmission of a particular trait either via the egg cell within the ovule or via the pollen as per experimental demand.

Fifthly, Mendel worked with a large number of plants, counted and subjected his findings to statistical analysis, and then compared his results with predictions based on mathematical models.

Finally, Mendel was a brilliant practical experimentalist. He could call and observe an optimum number of individuals from the limited space of the monastery garden.

In short, Mendel purposely set up a simplified ‘black and white’ experimental system and then successfully out how it worked.

Genetic Crosses With Guineapig

Mendel worked on pea plants but the application of his laws on animals was carried out by his successors.

Question 14. “Dispersal of animals is one of the driving forces of locomotion”— Judge the statement with the help of proper examples.
Answer:

Reasons And Motivations Behind Locomotion

Animals move for a variety of reasons, such as to find food, a mate, a suitable microhabitat, or to escape from predators.

Hunt For Food & Water:

Since animals can not prepare food, they are to perform locomotion to find out suitable area with plenty of food & water.

To seek shelter & escape from predators:

Locomotion increases the chances of survival of an organism by allowing the organism to seek shelter in favorable habitat, and to escape dangerous situations by avoiding predators.

Dispersal for mating or breeding:

Locomotion enables members of the species to disperse to find suitable mates to coordinate breeding activity and the survival of the young.

Search for a new & favorable environment:

Animals, especially birds, migrate from areas of low or decreasing resources to areas of high or increasing resources. The two primary resources being sought are food and nesting locations. Escaping from cold is also a factor.

Basic differences between movement and locomotion:

Though, in the case of animals, we often use the terms movement and locomotion, interchangeably, there are definite differences between the two.

Locomotion takes place at the organism level while movement can take place at any biological level from cellular to organisms.

In simple words, locomotion is movement from one place to another and involves shifting of the entire body parts (e.g. a person moving or running) and movement involves a change in shape, size, or direction of various body parts (e.g. shaking of your hands).

All kinds of locomotion are movements but all kinds of movements are not locomotion.

Movement is the motion that occurs in an organism or a body with or without the involvement of any change in the position or location of the organism or the body,

while locomotion is defined as the voluntary movement of an organism from one place to another either in search of food or shelter or mate or to escape from the predators.

Therefore, the differences between movement and locomotion may be summarized in a tabular form as below:

Question 15. Show the chemical constituents of chromosomes with the help of a table.
Answer:

Question 16. Compare DNA and RNA on the basis of the following features

1. Nature of carbohydrates
2. Nature of pyrimidine base

Answer:

Question 17. “Some phenotypes may have multiple genotypes and some other phenotypes may have a single genotype” justify the statement from the results derived from the dihybrid experiment in the case of the pea plant.
Answer:

Mendel’s Experiment And Laws For Dihybrid Cross

As a natural extension of the monohybrid cross, Mendel also designed experiments in which he examined two characters simultaneously. Such a cross, involving two pairs of contrasting traits, is called a dihybrid cross.

Experiment and Observation

Mendel-crossed pea plants that are heterozygous for two genes at the same time.

To construct such a dihybrid, he mated true-breeding plants grown from yellow round peas (YYRR) with true-breeding plants grown from green wrinkled peas (year).

From this cross, he obtained a dihybrid F₁ generation (YyRr) showing only the two dominant phenotypes, yellow and round. He then allowed these F₁ dihybrids to self-fertilize to produce the F₂ generation.

When Mendel counted the F₂ generation of one experiment, he found 315 yellow round, 101 yellow wrinkled, 108 round green, and 32 wrinkled green peas. There were, in fact, yellow wrinkled and green round recombinant phenotypes, providing evidence that some shuffling of alleles had taken place.

WBBSE Class 10 Life Science Previous Year Papers

Explanation

From the observed ratios, Mendel inferred the biological mechanism of shuffling the independent assortment of gene pairs during gamete formation.

Because the genes for peas’ color and for shape assort independently, Y can be with R or r in any gamete with equal probability.

Thus, the presence of a particular allele of one gene, say, the dominant Y for pea color, provides no information whatsoever about the alleles of the second gene.

That is, the allele for pea shape in Y carrying game could with equal likelihood be either R or r.

Each dihybrid of the F₁ generation can, therefore, make four kinds of gametes:

YR, Yr, yR, and yr. In a large number of gametes, the four kinds will appear in an almost perfect ratio of 1:1:1:1.

At fertilization then, in a mating of dihybrids, 4 different kinds of eggs can combine with any one of 4 different kinds of pollen, producing a total of 16 possible zygotes in the F₂ generation.

Once again, a Punnett square is a convenient way to visualize the process.

In fact, there are only nine different F₂ genotypes — YYRR, YYRr, YyRr, YyRR, yyRR, yyRr, YYrr, Yyrr, and yyrr — because the source of the alleles (egg or pollen) does not make any difference.

If we look at the combination of the traits determined by nine genotypes, we will see only four phenotypes— yellow round, yellow wrinkled, green round, and green wrinkled — observed in a ratio of 9 : 3 : 3: 1.

If, however, we look at just pea color or just pea shape, we can see that each trait is inherited in the 3: 1 ratio as predicted by Mendel’s law of segregation.

Punnet Square or Checker Board of Dihybrid cross

Model Question Paper Life Science Set 2 Group D Long Answer Type Questions

4. Write the 6 questions or their alternatives given below. Sightless candidates have to answer a question no. 4.1 A instead of question no 4.1. The marks allotted for each question is 5 (the division of marks is either 3+2, 2+3, or 5) :

Question 1. Draw a scientific diagram of a neuron and label the following parts :

1. Axon
2. Node of Ranvier
3. Dendron
4. Schwann Cell

Answer:

Draw a scientific diagram of the morphological structure of a eukaryotic chromosome and label the following parts—

1. Chromatid
2. Telomere
3. Centromere
4. Nucleolar Organizer

Answer:

Write the Functions of the following structural parts of a eukaryotic chromosome :

1. Centromere
2. Nucleolar Organizer
3. Chromatid
4. Telomere
5. Kinetochore

Answer:

Question 2. Compare the antagonistic functions of the following three pairs of hormones in the human body:

1. Insulin and Glucagon
2. Estrogen and Progesterone
3. FSH and LH

Explain the relationship between the hindbrain and the function of control of the breathing mechanism.

Antagonistic functions of insulin and glucagon :

1. Insulin increases cellular oxidation of glucose, stimulates the formation and storage of glycogen in the liver & muscles and inhibits the formation of glucose from non-carbohydrates. All these activities result in the maintenance of optimum levels of blood sugar.
Glucagon stimulates liver glycogen to undergo breakdown to be converted to glucose which increases blood sugar levels.

2. Insulin increases protein synthesis in the body. Glucagon hydrolyzes protein into amino acids.

3. Insulin decreases lipid & cholesterol levels in the blood. Glucagon stimulates the increase of fat & cholesterol level in the blood. Antagonistic functions of estrogen and progesterone :

1. Estrogen is secreted by ovaries prior to ovulation. Progesterone is secreted by ovaries after ovulation.

2. During pregnancy, estrogen induces enlargement of breasts and uterus. During pregnancy, progesterone reduces the contractility of the uterus and stimulates the growth of mammary glands.

Antagonistic features of FSH and LH :

1. FSH helps in the maturation of immature follicles into mature Graafian follicles. In the presence of FSH, LH ruptures Graafian follicles causing ovulation.

2. FSH stimulates the secretion of estrogen from Graafian follicles. LH stimulates the secretion of progesterone from the corpus luteum.

3. In males, FSH induces the development of seminiferous tubules and spermatogenesis. In males, LH stimulates the secretion of testosterone in the testis.

The breathing movements (ie., taking in oxygen and taking out carbon dioxide) are initiated in response to the ratio of carbon dioxide and oxygen within the body. High levels of C02 and low levels of 02 collectively influence the body to exhale C02 and take in 02.

This is regulated by the medulla oblongata of the hindbrain. In response to C02 and 02 ratios, medulla oblongata signals the heart and diaphragm accordingly to facilitate respiration.

The respiratory center of the medulla oblongata is concerned with controlling the depth and rate of respiration.

OR

The following physiological functions of a person wounded in an accident are found hampered. Write the name of the parts of the brain attached to those functions :

1. Speech
2. Hunger, thirst, and sleep
3. Posture and the equilibrium of the body
4. Movement of tongue and swallowing of food

Explain the mechanism of feedback control of the function of hormones with the help of a suitable example.

1. Speech — Cerebellum
2. Hunger, thirst, and sleep — Hypotha. lamus
3. Posture and the equilibrium of the body – Cerebellum

Movement of tongue and swallowing of food – Medulla oblongata

Answer:

Feedback control:

The secretion of most, if not all, hormones is regulated by some type of closed-loop control system known as a feedback mechanism

(because the amount released or secreted is sensed and that information is relayed back to the secretory cell by a variety of ways). Feedback control is mostly negative, rarely positive.

In a negative feedback control, the synthesis of a hormone slows down or halts when its level in the blood rises above the normal or threshold limit.

In the less common positive feedback mechanism, one hormone further stimulates the production of another hormone instead of diminishing it. Some examples of feedback control are given below.

Hypothalamus, in response to some external stimulus, produces a thyrotropin-releasing hormone for the secretion of thyrotropic hormone.

The thyrotropin-releasing hormone (TRH) stimulates the anterior pituitary lobe to secrete thyrotropic hormone. The latter in turn stimulates the thyroid gland to produce thyroxine.

If thyroxine is in excess, it exerts an influence on the hypothalamus and anterior pituitary lobe, which then secrete a lesser amount of releasing hormone and thyroid-stimulating hormone (TSH) respectively.

A rise in the TSH level in the blood may also exert a negative feedback effect on the hypothalamus and retard the secretion of TRH. This restores the normal blood-thyroxine level.

Sometimes, the accumulation of a biochemical increases its own production. For example, uterine contraction at the onset of labor stimulates the release of the hormone oxytocin, which intensifies uterine contractions.

The contractions further stimulate the production of oxytocin. This is a positive feedback control.

WBBSE Class 10 Life Science Previous Year Papers

Mode Of Transport Of Animal Hormones:

Most hormones are secreted into the general circulation to exert their effects on appropriate distant target tissues. Water-soluble hormone molecules circulate in watery blood plasma in a free state, (i.e. not attached to other molecules).

Steroid and thyroid hormones are less soluble in aqueous solution and over 90% circulate in blood as complexes bound to specific transport proteins like plasma globulins or albumin.

The transport proteins make the lipid-soluble hormones temporarily water-soluble and act as a ready reserve of hormones.

Functioning And Fate Of Hormones:

The chemical structure of a hormone enables it to combine with a receptor in the cells of its target. The receptor may be present on the plasma membrane of the cell or inside the cytoplasm or nucleus.

Only a hormone’s “target” cells, which have receptors for that hormone, will respond to its signal. When the hormone binds to its receptor, it forms a receptor-hormone complex.

This complex undergoes changes and enters into the nucleus of the target cell. Within the nucleus, it increases the synthesis of cell protein and takes part in the metabolic process of the cell.

All hormones diminish within the body at differing rates based on their chemical half-life. Once hormones have served their function on their target organs/tissues, these are destroyed.

These are either destroyed by the liver or the tissues of the target organs and excreted out of the body.

Role Of Hormones As Biochemical Messenger And Regulators:

Hormones are informational molecules that carry the message of metabolic change from the endocrine glands to the target cells or organizer

These are released into the extracellular fluid, where they are diffused into the bloodstream. The latter carries them from the site of production to the site of action.

Hormones stimulate or inhibit one or more Physiological processes for the welfare of the body.

Maintenance of the internal chemical environment of the body to a constant is called homeostasis. Hormones play a major role in maintaining homeostasis through their integrated action.

These also play a leading role in the chemical coordination of the living body. Thus hormones are known as chemical messengers.

Question 3. Prepare a list of the role of artificial plant hormones in agriculture and horticulture. What are the influences of Insulin hormone on the absorption and metabolism of glucose in the human body?
Answer:

Pancreas:

The pancreas is a large elongated gland located in the abdominal cavity just inferior and posterior to the stomach.

Hormonal Secretions:

The pancreas is considered to be a heterocrine or mixed gland as it contains both endocrine and exocrine tissue. The endocrine function consists primarily of the secretion of the two major hormones, insulin, and glucagon.

The endocrine cells of the pancreas are found in small groups throughout the pancreas called islets of Langer Within these islets are two major types of cells alpha and beta cells.

The alpha cells produce the hormone glucagon and the beta cells produce the hormone insulin.

Insulin:

It is an antidiabetogenic protein hormone that lowers blood sugar levels.

Functions:

Effect on carbohydrate metabolism:

Insulin increases the oxidation of glucose in the cells, stimulates the formation & storage of glycogen in the liver & muscle, inhibits the formation of glucose from non-carbohydrates in the liver, and increases the permeability of glucose through the cell membrane from blood to the cells.

All these activities result in the maintenance of optimum levels of blood sugar.

Effect on protein metabolism:

It increases protein synthesis in the body.

Effect on fat metabolism:

It also decreases lipid & cholesterol levels in the blood & prevents the formation of harmful ketone bodies in the liver.

Glucagon:

It is anti-insulin in nature. Together with insulin, it maintains a steady level of blood sugar in the body.

Functions:

Effect on carbohydrate metabolism: It stimulates liver glycogen to undergo breakdown to be converted into glucose which increases blood sugar levels.

WBBSE Class 10 Life Science Previous Year Papers

Effect on protein metabolism:

Glucagon hydrolyses protein into amino acids and increases nitrogen excretion through urine.

Effect on fat metabolism:

It stimulates the increase of fat & cholesterol level in the blood.

Gonads:

The gonads and ovaries in females and testes in males are responsible for producing the sex hormones of the body.

Testes:

The testes are a pair of ellipsoid Fun(Iami iniaIs of Lifi Sell no organs found in the scrotum of males that produce the androgen testosterone in males after the start of puberty.

Testosterone:

It has effects on many parts of the body, including the muscles, bones, sex organs (both primary & secondary), and hair follicles,

During puberty, testosterone controls the growth and development of the sex organs and secondary sex characteristics like muscular growth, breaking of voice, growth of hair on the face, chest, etc.

Ovaries:

The ovaries are a pair of almond-shaped glands located in the pelvic body cavity lateral and superior to the uterus in females. The ovaries produce the female sex hormones progesterone and estrogens. It also secretes another hormone called relaxin.

Estrogen:

Oestrogen stimulates the growth, development, and functional activities of primary and secondary sex organs in females during puberty.

It regulates the menstrual cycle and stimulates breast development.

It enhances the deposition of fat in the female body.

Progesterone:

In the presence of estrogen, progesterone stimulates the complete development of the primary & secondary female sex organizer

It is most active during ovulation & pregnancy. It helps the embedding of embryos in the uterus and stimulates the development & maintenance of pregnancy.

It also regulates the menstrual cycle & breast development at puberty.

Relaxin:

It is secreted from the uterus at the terminal stage of pregnancy. It facilitates the expulsion of the fetus from the uterus by causing relaxation of the pelvic ligaments.

OR

Write three differences between binocular vision with that of monocular vision on the following three aspects :

1. Formation of image
2. Field of vision
3. Depth

Write differences between inborn and acquired reflex actions on any two of the following aspects-

1. Nature
2. Condition
3. Prior experience
4. Neural pathway

Answer:

1. Applying the brake of a car at a red signal.
2. Sweating
3. opening the door on hearing the doorbell
4. looking left or right before crossing the road
5. knee jerk.

Mention the differences between conditioned and unconditioned reflex actions.

1. Conditioned reflex action
2. Unconditioned reflex action

Question 4. Show the alternation of generation in fern with the help of a word diagram.
Answer:
 

OR

Show the process of sexual reproduction in flowering plants with the help of a word diagram :
Answer: 

Question 5. Distinguish between Mitosis and Meiosis on the basis of the following three aspects-

1. Site of occurrence
2. Nature of division of chromosome
3. Number of cells produced

Explain the following phenomena related to meiotic cell division—

1. Separation of chromosome and chromatid
2. Crossing over

Answer:

Important features of meiosis

WBBSE Class 10 Life Science Previous Year Papers

1. Homologous Chromosomes:

These are chromosome pairs of approximately the same length and centromere position. Note that diploid cells have two sets or one pair of homologous chromosomes.

Out of the pair, one is inherited from the mother (maternal) and one from the father (paternal).

2. DNA replication:

Replication of DNA takes place during the interphase preceding meiosis. DNA replication generates sister chromatids from each chromosome. Sister chromatids are two identical copies of a chromatid that remain closely aligned.

Although DNA replication occurs in interphase, no longitudinal doubleness of chromosomes is visibly evident in Leptotene of Prophase I owing to contraction.

3. Synopsis:

The movement of chromosomes initiates in the zygotene stage and this movement results from an attracting force that brings the homologous pair of chromosomes together.

The chromosomes become shorter and thicker due to compaction.

When they come closer, homologous chromosomes pair and align at the equatorial plate of the cell for equal qualitative and quantitative distribution. The pairing takes place throughout the length.

This process of pairing is known as synapsis Pairing takes place not only between the homologous chromosomes but also between homologous regions of the chromosomes. Chromosome pairs undergoing synapsis have approximately the same length and centromere position.

Out of the pair, one is inherited from the mother (maternal) and one from the father (paternal). Non-sister chromatids belong to homologs.

WBBSE Class 10 Life Science Model Question Paper 2023

These are chromosome pairs having the same length, staining pattern, centromere position as well as the same characteristics of genes at particular loci.

A homologous pair of chromosomes consists of one chromosome from each parent and they are known as bivalents. Each chromosome of a bivalent is found to have two chromatids. Thus the four chromatids of a bivalent are together known as a tetrad.

4. Crossing Over:

The pachytene subphase of Prophase I is marked by a process called Crossing over that happens after both the homologous chromosomes in a pair join up to form a structure called a tetrad through synapsis.

Once a tetrad is formed, a portion of each homologous chromosome breaks off transversely and is re-attached to the same part of its homolog. Crossing over occurs between non-sister chromatids of homologous Sister Chiasma chromatids chromosomes.

This mixes up the traits that are found in each of the chromosomes, thus resulting in genetic recombination due to the exchange of segments between two non-sister chromatids belonging to a bivalent tetrad.

As a result of crossing over, X like structure is formed between the non-sister chromatids. The point of attachment at the crossing-over is called chiasma. It occurs during Diplotene.

The number of chiasmas may be one, two, or more depending on the length of the chromosomes.

5. Terminalization Of Chiasmata:

Chiasmata are generally pushed to the terminal ends of the chromosomes and this process is known as the terminalization of chiasmata (singular: chiasma).

The pair of homologous chromosomes begin to separate in the diplotene subphase of prophase I and chiasmata are fully terminalized in the diakinesis subphase of prophase.

Reduction in chromosome number:

During metaphase I, the tetrads line up on the plate. During anaphase I, the homologous chromosomes separate from one another i.e. one homologous chromosome with its two sister chromatids move to opposite poles.

Thus each pole has a haploid (n) number of chromosomes i.e. a reduction in chromosome number is achieved at this stage.

Telophase results in the formation of two haploid (n) daughter nuclei with each chromosome having two chromatids. For this reason, the first meiotic division is known as the reductional or heterotypic cell division.

7. Equational division:

Meiosis II or the second meiotic division has four phases similar to mitosis and is the second round of cell divisions during meiosis whereby the cells formed during Meiosis I divide again to form four haploid (n) gametes.

Between these two stages, the interkinesis or interphase is either quite short or skipped, unlike normal mitosis. Because of this, the S phase does not occur and so the DNA in these cells is not copied, making the resulting cells from this phase haploid.

OR

Distinguish between the cytokinesis of plant cells with the cytokinesis of animal cells on the basis of the following aspects :

1. Process
2. Time of initiation
3. Role of Golgi bodies

Answer:

The longest phase of mitosis:
Prophase

 The shortest phase of mitosis:
Anaphase

WBBSE Class 10 Life Science Model Question Paper 2023

Establish the interrelationships among genes, DNA, and chromosome.
Answer:

Inter-Relationship Among Chromosome, DNA, And Gene

The eukaryotic cells contain a membrane-bound nucleus which is designated as the ‘Director of the cell’.

The nucleus contains many thread-like, coiled, and elongated structures called chromatin fibers or chromatin reticulum, or nuclear reticulum.

The fibers of chromatin are uniformly distributed in the nucleoplasm. Chromatin fibers are observed only during the interphase or the preparatory phase of the cell cycle.

Just prior to cell division, chromatin fibers condense, become thick, and wrap up very tightly to form ribbon-like structures called chromosomes.

The number of particular species. They are usually found in pairs. Human beings have 23 pairs of chromosomes in each body cell.

The chromosome consists of a proteinaceous matrix and two spirally coiled chromonemata, each one of which contains a single DNA molecule.

The nucleus of an average human cell is only 6m in diameter, yet it contains 1.8 m of DNA which is distributed amongst the 23 pairs or 46 chromosomes each consisting of a single DNA molecule of about 40 mm.

Thus, Chromatin represents long-thin strands of the DNA-protein complex. It is unfolded and uncondensed form of DNA, while chromosomes are condensed DNA and protein.

The function of chromatin is to store DNA in the nucleus. Chromosomes are the bearers of hereditary instructions and regulate cellular processes.

DNA (or deoxyribonucleic acid) is the molecule that carries the genetic information in all cellular forms of life and some viruses.

It belongs to a class of molecules called nucleic acids, which are polynucleotides that is, long chains of nucleotides. Nucleotides are made up of a base, a sugar, and a phosphate. The four bases

chromosomes are constant in the nucleus for adenine (A), guanine (G), cytosine (C), and thymine (T) pair with each other (A with T and G with C).

WBBSE Class 10 Life Science Model Question Paper 2023

It is the order or sequence of these base pairs that provides the information needed for the growth and development of our bodies.

Specific parts of DNA carry the code for producing specific proteins which ultimately lead to the expression of different characters in an organism.

Each of these parts of DNA is termed a gene. Thus gene is the fundamental physical and functional unit of heredity that carries information from one generation to the next.

It is a segment of DNA, composed of a transcribed region and a regulatory sequence, that makes possible transcription. A series of genes are present in each DNA.

The coordinated interaction of two or more genes produces a given phenotypic trait. A complete set of chromosomal genes is inherited by the offspring as a unit from the parent.

Question 6. Predict what would be the hereditary results in the following cases :

1. One of the parents is thalassaemic and the other is a carrier of the thalassaemic gene.
2. Both of the parents are carriers of the thalassaemic gene.

What are the symptoms of Thalassaemia?
Answer:

OR

Tabulate three pairs of opposite traits regarding the seed of the pea plant as selected by Mendel. Show the process of sex determination in men with the help of a checkerboard.

Answer:

Model Question Paper Life Science Set 1 Group A

Answer to all questions is compulsory

1. Write the answer in complete sentences by choosing the correct answer for each question with the respective serial number.

Question 1. Select the correct pair-

1. Cerebrum-Maintenance of the balance of the body
2. Hypothalamus-Control of intelligence and emotion
3. Cerebellum-Control of body temperature
4. Medulla Oblongata-Control of heartbeat and swallowing of food

Answer: The correct Pair is- Medulla Oblongata—Control of heartbeat and swallowing of food.

Question 2. Identify which of the following statements is not true regarding Insulin-

1. Helps in the absorption of glucose from the blood into most of the somatic cells
2. Converts glucose into glycogen within liver and muscle cells
3. Helps in the conversion of fat and protein into glucose
4. Inhibits the conversion of protein and fat into glucose

Answer: The statement which is not true regarding Insulin is—Helps in the conversion of fat and protein into glucose.

Question 3. Match the words of Column-A with that of Column-B and select which of the following options is correct-

1. 1-1, 2-2,3-3
2. 1-2, 2-3, 3-1
3. 1-3, 2-1, 3-2
4. 1-2, 2-1, 3-3

Answer: 3. 1-3, 2-1, 3-2

Question 4 Determine from the answers given below in which phases of karyokinesis during mitotic cell division following two incidents happen- 1. Daughter chromosomes tend to move apart from each other towards their own poles Nuclear membrane and nucleolus disappear

1. 1. Prophase 2. Anaphase
2. 1. Anaphase 2. Prophase
3. 1. Telophase 2. Metaphase
4. 1. Metaphase 2. Telophase

Answer: 2. During mitotic cell division—

1. Daughter chromosomes tend to move apart from each other towards their own poles— Anaphase and
2. Nuclear membrane and nucleolus disappear—Prophase

Question 5. Select which of the following is the feature of cross-pollination-

1. Occurs within the same flower of the same plant
2. Agents are not required
3. Lesser chance of new characters being transmitted
4. More wastage of pollen grains

Answer: The feature of cross-pollination is— More wastage of pollen grains.

WBBSE Class 10 Life Science Model Question Paper 2023

Question 6. Determine the number of DNA molecules that get coiled to form each chromosome in a newly formed daughter cell produced by mitotic cell division in the human body-

1. 46
2. 1
3. 23
4. Numerous

Answer: 2. The number of DNA molecules that get coiled to form each chromosome in a newly formed daughter cell produced by mitotic cell division in the human body is 1.

Question 7. Identify the genotype of guineapig having black colour and rough hair-

1. BbRr, BBRr
2. bbRR, bbRr
3. BBrr, Bbrr
4. bbrr, bbRr

Answer: 1. The genotype guineapig having black colour and rough hair are—BbRr, BBRr.

Question 8. Decide which of the following two were selected by Mendel as recessive traits-

1. Colour of the flower-purple, the position of flower-axial
2. Length of stem-dwarf, a form of ripe seed-wrinkled
3. Form of the ripe seed-rund, the colour of seed- yellow
4. Position of flower-axial, length of stem- tall

Answer: 2. Length of stem—dwarf and Form of ripe seed—wrinkled were two recessive traits selected by Mendel.

Question 9. Asses from the following, the probable genotype of parents having a haemophilic son and normal daughter-

1. H || h, h
2. H || H, H
3. H || H, h
4. H || h, H

Answer: 3. The probable genotype of parents having a haemophilic son and normal daughter are- H || H, h

Question 10. Parthenium is an exotic species in our country. Other indigenous species cannot survive in such places where it grows. This establishes one of the postulates of Darwin’s theory. Identify the postulate-

1. Intraspecific struggle
2. Interspecific struggle
3. Struggle with environment
4. Origin of new species

Answer: 2. The correct postulate is Interspecific struggle.

Question 11. Miller and Urey, in their experiment, were able to synthesize some preliminary constituents necessary for the creation of life. Identify the ones which were amino acids among them-

1. Lactic acid, Acetic acid
2. Urea, Adenine
3. Glycine, Alanine
4. Formic acid, Acetic acid

Answer: The amino acids were—Glycine and Alanine.

WBBSE Class 10 Life Science Model Question Paper 2023

Question 12. Decide for which of the following purposes bees demonstrate waggle dance-

1. Search for reproductive mates
2. Inform other worker bees about the direction and the distance of the source of food from the bee hive
3. Selecting a place for the construction of the new bee hive
4. Avoid attack by probable enemy

Answer: 2. Bees demonstrate waggle dance to-^ Inform other worker bees about the direction and the distance of the source of food from the bee hive.

Question 13. Identify which of the following is the correct information related to biosphere reserve-

1. Conservation is promoted to local people and other biotic communities along with the conservation of the ecosystem
2. National Park and Sanctuary are not included within the biosphere reserve
3. The presence and participation of local people in the conservation of the ecosystem are not permissible
4. Its size is usually smaller than a Sanctuary

Answer: 1. The correct information related to biosphere reserve is- conservation is promoted to local people and other biotic communities along with the conservation of the ecosystem.

Question 14. Decide which of the following pair is not correct-

1. Poaching-Increase the endangeredness of gorilla
2. Exotic species-Lantana, Tilapia
3. Determination of hotspot-Number of endemic species and endangered species
4. Greenhouse gas-Eutrophication

Answer: The wrong pair is—A greenhouse gas— Eutrophication

Question 15. Decide which of the following Project Tiger is located within our state-

1. Bandipur
2. Simlipal
3. Sunderbans
4. Kanha

Answer: Sunderbans Project Tiger is located within our state.

WBBSE Class 10 Life Science Model Question Paper 2023

Model Question Paper Life Science Set 1 Group B

2. Answer any 21 questions out of the 26 questions given below instructed.

Fill in the blanks with proper words in the following sentences (any five):

Question 1. Acharya Jagadish Chandra Bose proved the property of __________ by sending electrical impulses in Mimosa and Desmodium plants.
Answer: Sensitivity

Question 2. If gametes in humans were produced by mitosis instead of meiosis then the number of autosomes in a somatic cell of an offspring would have been __________.
Answer: 88

Question 3. A disease in the human population caused by a recessive gene located in the ‘X’ Chromosome is __________.
Answer: Haemophilia

Question 4. The hoof of modern horse is the transformation of the digit number __________ of their ancestors.
Answer: 3

Question 5. At the __________ phase of the nitrogen cycle, ammonia is converted into nitrite and nitrate by the action of some bacteria.
Answer: Nitrification

Question 6. To produce the bottled cold drinks widely sold in the market, a lot of __________ water is wasted.
Answer: 2.6 Fresh

Decide whether the following statements are true or false (any five):

Question 7. Tropic movement is the movement of growth in plants.
Answer: True

WBBSE Class 10 Life Science Model Question Paper 2023

Question 8. Crossing over takes place during mitotic cell division.
Answer: False

Question 9. Mendel used the term gene while describing his experiments related with heredity.
Answer: False

Question 10. The leaf of the Cactus is modified into the spine for the reduction of the rate of transpiration.
Answer: True

Question 11. Rhododendron is an endangered plant species conserved in the Eastern Himalaya hotspot.
Answer: True

Question 12. Choroid helps in the accommodation of the eye by changing the curvature and shape of the lens.
Answer: False

Match the words in Column-A with those which are most appropriate in Column-B and re-write the correct pair mentioning the serial no. of both Columns (any five):

Answer: 13. D, 14. G, 15. A, 16. E, 17. B, 18. C

Answer in a single word or in a single sentence (any six):

Question 19. Choose the odd one and write it: Cerebrum, Hypothalamus, Pons, Thalamus
Answer: Pons

Class 10 Life Science Model Paper WBBSE 2023

Question 20. Where the Schwann cells are located?
Answer: Schwann cells are found in close contact with axons in the peripheral nerves.

Question 21. A pair of related terms is given below. On the basis of the relationship in the first pair write the suitable word in the gap of the second pair. Mitosis: Radicle :: Meiosis: __________.
Answer: Spore mother cell

Question 22. Which law did Mendel conclude from his dihybrid cross experiment?
Answer: Law of Independent Assortment

Question 23. Give an example of a variation found among healthy persons which is transmitted through generations.
Answer: Rolling tongue and non-rolling tongue

Question 24. How do Chimpanzees break open the hard shells for eating the nuts?
Answer: Chimpanzees put the nut on the flat surface of a hard stone and use another stone or tough piece of wooden branch as a hammer to break the shell of nut.

Question 25. Among the following four terms, one includes the other three. Find it out and write it: SPM, Air Pollution, Greenhouse gas, Lung disease.
Answer: Air pollution

Question 26. Name the practice which jointly the local peoples and forest department maintain for the reclamation of a forest.
Answer: Joint Forest Management (JFM)

Model Question Paper Life Science Set 1 Group C

3. Answer any 12 questions in 2-3 sentences out of the 17 questions given below.

Question 1. Distinguish between the functions of hormones and the nervous system on the following parameters:

1. Nature of function
2. Pace of function
3. A time span of function
4. Fate

Answer:

Differences between the endocrine system and nervous system

Question 2. ‘A person can see distant objects distinctly but is unable to visualize near objects in a perfect way’-Predict what would be the probable cause and suggest the corrective measure for such a problem.

Answer:

Defects Of Visions And Corrective Measures

A person with normal eyes can, by virtue of accommodation, see clearly all objects that are at a distance greater than about 25 cm from the eye.

If due to certain abnormalities the eye is unable to accommodate itself to various distances, then the eye is said to be defective.

Some common defects of the eye are—

1. Myopia:

Near-sightedness, also called myopia is the common name for impaired vision in which a person sees near objects clearly while distant objects appear blurred.

In such a defective eye, the image of a distant object is formed in front of the retina and not at the retina itself. Consequently, a nearsighted person cannot focus clearly on an object farther away than the far point of the defective eye.

Class 10 Life Science Model Paper WBBSE 2023

Causes:

This defect arises because the power of the eye is too great due to the decrease in the focal length of the crystalline lens.

This may arise due to either-

Excessive curvature of the cornea, or

Elongation of the eyeball.

Correction:

This defect can be corrected by using a concave (diverging) lens. A concave lens of appropriate minus (-) power or focal length is able to bring the image of the object back to the retina itself.

2. Hyperopia or hypermetropia:

Far-sightedness, also called hyperopia or hypermetropia, is the common name for a defect in vision in which a person sees near objects with blurred vision, while distant objects appear in sharp focus.

In this case, the image is formed behind the retina.

Causes:

This defect arises because either

the focal length of the eye lens is too great, or

the eyeball becomes too short so that light rays from the nearby object cannot be brought to focus on the retina to give a distinct image.

Correction:

This defect can be corrected by using a convex (converging) lens of appropriate focal length. Eyeglasses with converging lenses supply the additional focusing plus (+) power required for forming the image on the retina.

3. Presbyopia:

Presbyopia is a progressive form of farsightedness that affects most people by their early 40s. The power of accommodation of the eye decreases with aging.

Most people find that the near point gradually recedes.

Causes:

It arises due to the gradual weakening of the ciliary muscles and diminishing flexibility of the crystalline lens.

Correction:

Simple reading eyeglasses with convex lenses correct most cases of presbyopia.

Sometimes, a person may suffer from both myopia and hypermetropia. Such people often require bifocal lenses.

In the bifocal lens, the upper portion of the bifocal lens is a concave lens, used for distant vision. The lower part of the bi-focal lens is a convex lens, used for reading purposes.

Class 10 Life Science Model Paper WBBSE 2023

Cataract:

Generally, this defect can be found in aged or old people. Persons with this defect get blurred vision which sometimes even lead to total blindness.

The reason for this defect is that the lens loses its transparency and become opaque due to the deposition of protein material and calcium mineral in the lens.

This opaque condition of the lens does not allow the light rays from an object to pass through the lens. This defect can be rectified by surgically removing the lens and it has to be replaced by a highly convex lens.

Before intraocular lenses (lOLs) were developed, people had to wear very thick eyeglasses or special contact lenses to be able to see after cataract surgery.

Now, with cataract lens replacement by phacoemulsification or phaco surgery, several types of IOL implants are available to help people enjoy improved vision.

Question 3. Analyze the role of synthetic plant hormones in increasing production and solving the problem of weeds in agriculture.

Answer:

Role of synthetic hormones:

Synthetic hormones are successfully used in agriculture and horticulture. The roles of these synthetic hormones are mentioned below.

1. Developing new plants from stem cuttings:

Cuttings are used for artificial vegetative propagation of different plants like roses, Hibiscus, marigold, Chrysanthemum, ‘etc. After cutting the twigs from a mother plant, a solution of synthetic auxin or auxin powder is applied at the cut end.

Then, these cuttings are planted in moistened soil. By the action of this hormone, adventitious roots grow from the cut end and the cutting grows as an individual daughter plant.

2. Preventing shedding of immature fruits:

Sometimes, immature fruits shed off from the plant if these plants are sprayed with synthetic auxin solution for a few times during ear y developmental phase of the fruits, the rate of immature shedding declines sharply.

Horticulturists spray auxin solution on mango, litchi, grapes, banana, and several other fruit plants to prevent immature shedding of fruits. Synthetic gibberellin and synthetic cytokinin are also effective.

Class 10 Life Science Model Question Paper With Answers

3. Destroying weeds:

Weeds growing in crop fields share water and nutrients with agricultural crops. This affects the quality of production. Scientists have revealed that the application of certain phytohormones destroys dicotyledonous herbs and shrubs.

Application of a synthetic auxin named 2, 4-D effectively kills dicotyledonous weeds from monocot crop (paddy, wheat, etc.) fields.

4. Production of parthenocarpic fruits:

Fruits, produced from the ovary without fertilization do not contain seeds and become larger. These are called parthenocarpic fruits. A treatment of auxin solution before the maturation of flowers triggers the development of the ovary.

As a result, seedless fruits are produced before fertilization. Synthetic auxin is successfully applied on the plants of guava, grapes, banana, watermelon, etc. to produce seedless fruits.

Synthetic gibberellin is comparatively more effective on tomato plants. These two synthetic phytohormones are used to produce parthenocarpic fruits.

Question 4. LH and ICSH control the secretion of hormones of the reproductive glands of the human body’-judge the validity of the statement.

Answer:

In the female body, Lutenising Hormone (LH) stimulates ovule secretion and the formation of corpus luteum from the Graafian follicle. It also helps in the secretion of progesterone hormone from the corpus luteum.

In the male body, Interstitial Cell Stimulating Hormone (ICSH) stimulates the interstitial cells of Leydig of testes to secrete testosterone.

Question 5. How can you distinguish between the mitosis of plant cells and that of animal cells on the basis of the formation of spindle fiber and the process of cytokinesis?

Answer:

Distinguish between the mitosis of plant cells and that of animal cells are-

Question 6. Establish the relationship between the formation of malignant tumors in the human body with the loss of control in the cell cycle.

Answer:

The relationship between the formation of malignant tumors in the human body with the loss of control in the cell cycle

Checkpoints prevent uncontrolled cellular growth and thereby cancer. It mainly checks the genetic as well as physical integrity, if they found any genetic defect, they immediately arrest the cell from going to cellular division.

All cells will be checked in three stages. But if the genes which are responsible for the synthesis of checkpoints are mutated then they lost their control over cell division and the cells are going to divide in an uncontrolled way.

This uncontrolled cellular division gives rise to a cellular lump, called a tumor.

Class 10 Life Science Model Question Paper With Answers

Question 7. ‘The adventitious leaf bud plays a significant role in natural vegetative propagation of plant’-Evaluate the validity of the statement with a proper example.
Answer:

Natural Vegetative Propagation

Different plant parts are variously modified for vegetative propagation.

Natural vegetative propagation in higher plants:

Root:

The storage root acts as a structure for perpetuation, they help the organism to survive during unfavorable conditions and also germinate to produce a new plant body. e.g. Root tuber of sweet potato, Dahlia.

Stem:

The stem may also act as a parenting organ, which can grow and propagate on land and water and when separated, produces a new plant body. e.g. Offset of Water Hyacinth.

The propagation of stem in terrestrial plants is brought about by runner and stolon, e.g. There are various artificial methods like strawberries and grasses. Rhizomes are cutting, grafting, micropropagation, etc.

underground horizontally growing stems having nodes, internodes, and axillary buds. Branches grow from these buds.
Tuber is a modified underground stem tip.

The eyes or buds present on the tuber grow into new plants. The bulb is a modified shoot that has short stems and apical & axillary buds that grow to form shoots.

Leaf:

The leaf may become a propagating organ with the presence of storage food in the lamina and adventitious buds present at the edge of the lamina may germinate to produce a new plant body. e.g. Leaf of Bryophyllum.

Question 8. Show with the help of a cross, who is more important among parents in determining the sex of their offspring.
Answer: 

Sex Determination In Human

The term sex refers to sexual phenotype. Most organisms have only two sexual phenotypes—male and female.

We, normally, define the sex of an individual organism in reference to its phenotype. The mechanism by which sex is established is termed sex determination.

Sometimes an individual organism has chromosomes that are normally associated with one sex. For example, the cells of female humans normally have two X chromosomes, and the cells of males have one X chromosome and one Y chromosome.

Class 10 Life Science Model Question Paper With Answers

Sex Determination In Humans:

In humans, Drosophila, and many other species, the cells of males and females have the same number of chromosomes, but the cells of males have a single X chromosome and a smaller sex chromosome, the Y chromosome.

The Y chromosome is not Y shaped as is commonly assumed but is acrocentric. In this type of sex determination system, the male (44A + XY) is heterogametic because half of the male gametes have an X chromosome and the other half have a Y chromosome.

The female (44A + XX) is homogametic because all the eggs contain a single X chromosome. Fertilization of an egg (always X-bearing) with an X-bearing sperm produces female offspring (XX), but a Y-bearing sperm produces male offspring (XY).

The Total Chromosomes In Humans, In Each Body Cell, Can Be Represented As-

Female chromosomes – 44+XX where 44 are the autosomes and XX chromosomes are the sex chromosomes.

Male chromosomes – 44+XY where 44 are the autosomes and XY chromosomes are the sex chromosomes.

Although the X and Y chromosomes are not homologous, they pair and segregate into different cells in meiosis.

This is because of the fact that these chromosomes are homologous in small regions, called the pseudoautosomal region, in which they carry the same genes. In both types, the human X and Y chromosomes contain pseudoautosomal regions.

In humans and other placental mammals, maleness is due to a dominant effect of the Y chromosome. This is evidenced by the study of individuals with an abnormal number of sex chromosomes or aneuploidy.

XO persons (Turner syndrome) develop as females and XXY persons (Klinefelter syndrome) develop as males.

The dominant effect of the Y chromosome is exhibited early in development when it directs the primordial gonads to develop into testes.

Once the testes are formed, they secrete the hormone testosterone, which stimulates the development of male secondary sexual characteristics.

It is now known that the testis-determining factor (TDF) is the product of a gene called SRY (Sex-determining region Y), which is located outside the pseudoautosomal region in the short arm of the Y chromosome.

When fertilization occurs, the zygote (the initial cell from which a fetus grows) always inherits one of the mother’s X chromosomes, and either an X or a Y from the father, depending on which chromosome the fertilizing sperm cell happened to inherit.

One could say, then, that the father or, at least, his sperm determines the sex of the child.

The generally accepted theory is that males determine the sex because males can donate either an X chromosome or Y chromosome, while females can only donate an X chromosome to their offspring, making their contribution constant and the male’s contribution.

The variable, which under normal circumstances, determines the offspring’s genetic sex (at least, in humans Moreover, genetically there is a 50% chance of having a boy and a 50% chance of having a girl, as is found out from.

But there are actually slightly more boys born every year than girls.

It’s unclear why this is the case, but some research points out that more female fetuses die during pregnancy than male.

The Y chromosome contains all the directions that make the human zygote develop into a male. It is a relatively small chromosome with about 30 genes.

WBBSE Class 10 Life Science Previous Year Papers

In comparison, the X chromosome has between 800 and 900 genes.

With its limited number, the Y chromosome focuses primarily on male traits. It contains the all-important SRY gene, which instructs the embryo to develop male traits such as testicles.

Another gene unique to the Y chromosome is USP9Y, which contributes to sperm production.

Question 9. ‘Different genotypes produce the same phenotype’-Justify the statement in the form of a table by taking an example from the result of the dihybrid cross of the Pea plant.
Answer:

The seven pairs of characters as chosen by Mendel

Mating between individuals that differ in only one trait, such as seed color or stem length is known as a monohybrid cross.

In each monohybrid cross, one parent carries one form of the trait, and the other parent carries an alternative form of the same trait. Mendel selected seven such traits to study the monohybrid breeding experiment.

Each trait had two easily distinguishable, alternative appearances (phenotypes).

These are—

Experiment And Observation

Mendel carried out a series of monohybrid crosses.

For example, in the spring of 1854, he planted pure-breeding green peas and pure-breeding yellow peas and allowed them to grow into the parental (P) generation.

Later that spring when the plants had flowered, he dusted the female stigma of green-pea plant flowers with pollens from yellow-pea plants. He also performed the reciprocal cross between the female yellow pea plant and the male green pea plant.

In the fall (autumn), when he collected and separately analyzed the progeny peas he found that in both cases, the peas were all yellow.

The yellow peas, the progeny of the P generation, were the beginning of what we now call the first filial (Fx) generation. Mendel planted them and allowed the F₁ plants to self-fertilize.

He then harvested and counted the peas of the resulting second filial (F₂) generation, the progeny of the Fx generation.

Among the progeny of one series of F₁ self-fertilization, there were 6022 yellow and 2001 green peas, an almost perfect ratio of 3 yellow to 1 green.

The results of reciprocal crosses produced a similar ratio. Monohybrid crosses involving other traits (such as long and short stem length) also showed similar results.

Question 10. Give your opinion about probable suggestions which can be given to a pair of contenders before marriage in order to prevent the spread of a genetic disease from the society already known to you.
Answer:

Genetic Counselling

Thalassemia mutations and various abnormal hemoglobins interact to produce a wide range of disorders of varying degrees of severity.

Hemoglobin disorders are the most common worldwide inherited conditions. They are common in populations of tropical Africa, Asia, and the Mediterranean region and are spreading by migration throughout the world.

WBBSE Class 10 Life Science Previous Year Papers

A stem cell transplant is the only treatment that can cure thalassemia. But only a small number of people who have severe thalassemias are able to find a good donor match and have the risky as well as expensive procedure.

Genetic counseling, therefore, plays the most important part in thalassemia prevention programs considering the diversity and severity of the problem.

Genetic counseling is defined as the process by which patients or relatives at risk of a disorder that may be hereditary are advised of the consequences of the disorder and the probability of developing and transmitting it and the ways in which this may be prevented.

It Involves The Followings:

A correct diagnosis using genetic tests after evaluation of family history and medical records Explanation of the nature of disorder & the treatment available Estimation of genetic risk for parents and family members.

Communication of genetic risks and the options for avoiding them to avoid undue complications in an unbiased manner Support in making the right decision Accessibility for long-term contact and counseling.

Premarital screening for thalassemia and sickle cell should be made mandatory to decrease at-risk marriages.

The objective is to make people aware of the consequences of thalassemia on health and socio-economics so that they voluntarily ask for screening, and prevention and change their reproductive plans when a possible risk is found.

If the risk is found before marriage, the options are to remain single, not to marry another carrier or to marry irrespective of carrier status.

If the risk is found after marriage, the options are to separate and find a non-carrier partner, to have a few or no children, selective termination of pregnancy, or to take a chance and have children as usual.

There are therefore challenges involved in genetic counseling because all the available choices involve difficult moral and social problems and in most cases, there appears to be no right answer.

But on the other hand, once people understand the risk, they can not escape from making a choice even if the decision ‘not to choose’ is a choice.

Question 11. A good number of Tilapia fishes are released in a pond having only different indigenous fish species grown naturally. Think and write which types of the struggle for existence Tilapia fishes have to face in order to survive.
Answer:

The Tilapia fishes have to face three types of struggle for existence in the pond where they are released.

These are

Intra-specific struggle:

It is the struggle among all the newly released tilapia fishes for food and habitat

Inter-specific struggle:

It is the struggle between Tilapia and other naturally grown indigenous fishes for food and habitat.

Struggle with the environment:

The Tilapia fishes struggle with the condition of the pond where they are newly released. Water content, Temperature of water, amount of dissolved oxygen, poisonous substances present in that pond water, etc. are the obstacles that they struggle, to overcome.

Question 12. Prepare a list of roles air sacs of pigeons play to fly in the air.

Answer:

Importance of air sacs in pigeons:

Pigeon is a primary volant animal. It has nine non-vascular and non-muscular air sacs, emerging from the bronchioles, typically helping in flight.

The importance of air sacs in pigeons is-

1. During the flight, the pigeon needs extra energy, the production of which requires additional oxygen. Oxygen concentration near the ground level is higher than that in the high sky. Before the flight, pigeons fill the air sacs with this oxygen-rich air. When in the high sky, these air sacs supply oxygen-rich air to the lungs.
2. The air-filled sacs decrease the specific gravity of the body of the pigeon, which is also beneficial for flight.

Importance of air bladder in rohu fish:

1. The air bladder in rohu fish helps it to move up and down in the water. The air bladder changes the buoyancy of the body and thus, assists it to go at different depths of water.
2. The red gland in the anterior chamber of the air bladder fills gas in it to reduce the specific gravity of the body and thus helps the fish to move upward in water. On the other hand, the rete mirabile of the posterior chamber absorbs the gas to increase the specific gravity of the body of fish and helps the fish to go deeper in the water.

WBBSE Class 10 Life Science Previous Year Papers

Question 13.

1. Structure and function
2. Indicating the nature of evolution

Based on the above two features establish the concept of the analogous organ with the help of a proper example.

Answer:

Analogous organs are those organs that are different in structure and origin but perform similar functions.

Analogous organs indicate convergent evolution. For example, the wings of birds, wings of insects, and patagium of bats perform the same function, i.e., they help the organisms to fly but they have different structures.

Wings of birds are modifications of forelimbs, wings of insects are an outgrowth of insects’ exoskeleton and the patagium of the bat is actually a fold of skin between forelimbs and hind limbs.

All these structures point toward the fact that in the same environment, different structures may evolve to perform the same function. This further points toward convergent evolution.

Question 14. Relate the following phenomena with the trend of disturbance in the Nitrogen Cycle resulting from different human activities:

1. Global Warming
2. Acidification of soil and water of river and lake.

Answer:

Global warming:

Around 40% increase of N₂0 (nitrous oxide) in the environment is due to human activities. N₂0 is released in the environment as a result of the combustion of fossil fuel and the overuse of nitrogen-rich fertilizers.

This N₂O is a greenhouse gas that absorbs infrared radiation or heat of the sunlight which reflects on the earth and facilitates the greenhouse effect and global warming.

Acidification of soil and water of rivers and lakes:

Oxides of nitrogen get dissolved in rainwater and form nitric acid, which is a major component of acid rain. Acid rain destroys aquatic plants and animals and disrupts the ecological equilibrium of rivers lakes, ponds, etc.

Apart from this, the destruction of forest resources and harm to architecture and monuments take place due to acid rain.

Question 15. Hilsa, Bee, Penguin, Rauwolfia- Assess which are the causes of the endangeredness of the above-mentioned organisms.

Answer:

Hilsa:

Facing threat due to overexploitation of this population.

Bee:

Loss of habitat, and excessive use of cell phones that project electromagnetic waves, damage the ability of bees to return to their colony. In that way, it destroys the navigation system of bees.

Penguin:

Penguin faces threats from geological events like a volcanic eruption, pollution, climatic changes, and severe weather.

Rauwolfia:

This type of medicinal plant faces threats due to overexploitation.

WBBSE Class 10 Life Science Model Question Paper 2023

Question 16. Discuss any two roles of the People’s Biodiversity Register (PBR) in conserving local biodiversity.

Answer:

Joint Forest Management or JFM:

The proper management of biodiversity by the joint action of local people and the forest department of the state government which is approved by the Indian government, is known as Joint Forest Management or JFM.

Role of JFM in the conservation of biodiversity:

In the conservation of biodiversity, the role of JFM is as follows

1. The committee, formed by the local residents of any forest area, maintains coordination with the state forest department regarding the conservation of the biodiversity of that forest.
2. The JFM committee members keep a vigil on the forest area along with their daily activities regarding forest fire, encroaching of the forest area, poaching, felling of trees, etc., and report it immediately to the forest department in case of any adversity.

People’s Biodiversity Register or PBR:

It is a ready reference prepared by the central forest department in accordance with the Biodiversity Law 2002. This document contains detailed information on the availability of local biodiversity, their habits and habitats, other behaviors, their ecological importance, etc.

Role of PBR in the conservation of biodiversity:

In the conservation of biodiversity, the role of PBR is as follows-

1. PBR provides correct information about the biodiversity of the local areas so that, the rare and endangered animals and plants can be conserved properly.
2. It makes people aware of the importance of the conservation of biodiversity and involves them directly with the conservation project.
3. PBR makes the people aware of the sustainable utilization of forest resources and helps in the economic development of the area.
4. PBR provides correct geographic information about conserved areas.
5. Identify endangered species and collect information about the conservation of these species.
6. Gather knowledge about local species and their cultural and social value.
7. Gives an idea about the obtained product from species and their value along with their collection and transport procedure.

Question 17. Tabulate any of the four activities which are prohibited in a sanctuary following the provisions of the Wildlife Act.

Answer:

The four activities, that are prohibited in the sanctuary are-

1. Human intervention in the sanctuary is strictly prohibited.
2. Any human activity which disturbs wild animals is prohibited.
3. Poaching, hunting, and fishing are strictly prohibited.
4. Felling of trees is completely prohibited.

WBBSE Class 10 Life Science Model Question Paper 2023

Model Question Paper Life Science Set 1 Group D

Answer 6 questions or their alternatives given below.

Question 1. Draw a neat diagram of the vertical section of the eyeball of the human eye and label the following parts:

In order to be able to learn about and/or solve problems presented to them by their environment, humans must be able to take in information from that environment;

process it quickly and accurately; decide when, how, and what action to take; and then perform or execute that action. In order to receive information from the environment we are equipped with sense organizer

A sense organ is a specialized bodily structure that receives or is sensitive to internal or external stimuli. It is composed of receptors. Human sense organs are the eyes, ears, tongue, skin, and nose.

The Eye As A Sense Organ-Its Location, Structure, And Function

The sense organ consisting of photosensitive receptors (the rod & cone cells) and which helps us in vision is known as the eye.

The two eyes are located in the deep cavities of the skull called orbits on the frontal part of the cranium.

The structure of different parts of the human eye and their corresponding functions are given below:

Or,
Explain the importance of the following parts of an eukaryotic chromosome:

1. Centromere
2. Telomere

Analyze the role of cell divisions in controlling growth, reproduction, and repair in an organism.

Answer:

The physical structure of eukaryotic chromosomes:

The morphological features of a chromosome appear ‘most distinctly under a microscope during the metaphase stage of cell division. From this study, we can find 5 parts of a chromosome. These parts are described below.

1. Chromatids:

In a metaphase chromosome, two identical and longitudinal strands are seen. These are chromatids. Two chromatids of the same chromosome are called sister chromatids, which remain attached to a constricted region or centromere.

Each chromatid carries one or a few very fine filaments along its length. These are called chromonemata (singular—chromonema). Each chromonema is composed of a longitudinally arranged coiled DNA.

Along each chromonema, several spherical linearly arranged bead-like structures are seen, which are called chronometers.

2. Primary constriction and centromere:

Each chromosome has a distinct constricted region at which the sister chromatids remain attached to each other. This is known as primary constriction.

At e primary constriction, a round plate-like and dense heterochromatin structure is seen, which is called the centromere. The centromere has a few adhesive points, called kinetochores, which attach to the spindle fibers during metaphase.

The DNA present in the .centromere is genetically inactive in nature.

WBBSE Class 10 Life Science Model Question Paper 2023

3. Secondary constriction:

Other than primary constriction, there are one or a few constricted regions in the chromosomes. These are called secondary constrictions. Generally, the nucleolus is seen affixed to the secondary constriction.

During the telophase of cell division, this region helps to reorganize the nucleolus. Therefore, secondary constriction is also known as nucleolar organizer region or NOR.

4. Satellite:

In a few chromosomes, a bulb-shaped terminal portion is seen beyond the secondary constriction. This is called a satellite or SAT body. The chromosomes with SAT body are called SAT chromosomes.

5. Telomere:

The terminal portions of a chromosome are called telomeres. These are genetically inactive regions of a chromosome. During interphase, telomeres help in DNA replication.

It also prevents the joining of a chromosome with another and controls the ageing and death of a cell.

Significance of cell cycle:

The significance of the cell cycle is mentioned below-

1. Controlling cell division:

Certain points of the cell cycle control cell division. These are known as checkpoints. In case of any functional disruption at any of those points, the cell division process becomes uncontrolled, which may lead to tumor formation.

Tumors are of two types-benign tumors and malignant tumors. Benign tumors are harmless but malignant tumor cells invade other tissues through blood or lymph and form tumors there.

This phenomenon is called metastasis, which is a characteristic feature of cancer cells.

2. Normal growth and wound healing:

Cell division helps an organism to grow in size. It also assists in wound healing.

Importance of cell division:

The importance of the cell cycle is given below.

1. Growth:

The number of cells in an organism increases due to cell division. The daughter cells produced by this process also grow in size. Therefore, the growth of any organism depends directly upon cell division.

2. Reproduction:

Amitosis, mitosis, and meiosis help in different types of reproduction processes. Simple unicellular organisms like Amoeba, reproduce by the amitosis process. Mitosis helps in asexual and vegetative reproduction in animals and plants.

By meiotic division, gametes and spores are formed. Therefore, meiosis helps in sexual and asexual reproduction.

3. Wound healing:

Mitosis helps in the repair of wounds and the regeneration of organs in plants and animals.

4. Transfer of genetic characters:

By cell division, the characters of the mother cell are transferred to the daughter cells. From a broader perspective, the newer characters in daughter cells help in adaptation and evolution.

Question 3. Tabulate three pairs of dominant-recessive traits of pea plants as selected by Mendel. State the first law of Mendel as deducted from the experiment of the Monohybrid cross.

Answer :

Three pairs of dominant-recessive traits of pea plant as selected by Mendel

Or,
A color-blind female married a normal male. Judge the probability of color blindness among their children in the first filial generation. Show with the help of a cross how the first law of Mendel deviates in the case of the Four-O clock plant in F2 generation.

Answer:

Inheritance of color blindness:

Sex-Linked Inheritance

The gene is located in the sex chromosome. Sex chromosomes may be X or Y chromosomes.

But, as the Y chromosome is more or less genetically inert (functionless or inactive) with respect to the X chromosome, we generally consider sex-linked genes and X-linked genes to be synonymous.

Some X-linked recessive characteristics are- color blindness, hemophilia, etc.

Haemophilia

Haemophilia is a rare disorder in which blood doesn’t clot normally because it lacks sufficient blood-clotting proteins (clotting factors) causing the sufferer to bleed severely even from a slight injury.

Symptoms:

The major signs and symptoms of hemophilia are excessive bleeding and easy bruising.

Children who have mild hemophilia may not have signs unless they have excessive bleeding from a dental procedure, an accident, or surgery,

Bleeding can occur on the body’s surface (external bleeding) or inside the body (internal bleeding)— blood in stool, blood in urine, bruising, nose bleeds, heavy or prolonged periods, etc.

Swollen joints and pain in the joints.

Causes:

A defect in one of the genes that determine how the body makes blood clotting factor VIII or IX causes hemophilia. These genes are located on the X chromosomes.

Chromosomes come in pairs. The two types of hemophilia are factor VIII deficiency (hemophilia A) and factor IX deficiency (hemophilia B, or Christmas disease).

The most common type of hemophilia is called hemophilia A in which the person does not have enough clotting factor VIII (factor eight).

A less common type is hemophilia B in which a person does not have enough clotting factor IX (factor nine). Females have two X chromosomes, while males have one X and one Y chromosome.

Only the X chromosome carries the genes related to clotting factors. Haemophilia usually occurs more in males than in females. About 1 in 5,000 males are born with hemophilia each year.

A male who has a hemophilia gene on his X chromosome will have hemophilia. When a female has a hemophilia gene on only one of her X chromosomes, she does not have symptoms of hemophilia, since at least one of the X- chromosomes have a factor VIII or IX gene that works to produce normal or near normal levels of factors.

Class 10 Life Science Model Paper WBBSE 2023

However, some women who carry this gene may have a bleeding tendency. They are called ‘symptomatic carriers’. They are recognized as having mild hemophilia.

In very rare cases, some women have particularly low factor levels causing them to have moderate or severe hemophilia. In such cases, both X chromosomes are affected or one is affected and the other is missing or inactive.

Inheritance Of Haemophilia:

In this example, the mother is a carrier of the hemophilia gene and the father does not have hemophilia There is a 50% chance that each son will have hemophilia.

There is a 50% chance that each daughter will be a carrier of the hemophilia gene.

In this example, the father has hemophilia and the mother does not carry the hemophilia gene:

None of the sons will have hemophilia All daughters will carry the hemophilia gene In this example, the father does not have hemophilia and the mother does not carry the hemophilia gene

None of the children (either daughters or sons) will have hemophilia or carry the gene.

A Royal Disease:

Haemophilia is sometimes referred to as “the royal disease,” because it affected the royal families of England, Germany, Russia, and Spain in the 19th and 20th centuries.

Queen Victoria of England, who ruled from 1837-1901, is believed to have been the carrier of hemophilia B, or factor IX deficiency. She passed the trait on to three of her nine children.

Her son Leopold died of a hemorrhage after a fall when he was 30. Her daughters Alice and Beatrice passed it on to several of their children.

Alice’s daughter Alix married Tsar Nicholas of Russia, whose son Alexei had hemophilia.

Their family’s entanglement with Rasputin, the Russian mystic, and their deaths during the Bolshevik Revolution have been chronicled in several books and films.

Haemophilia was carried through various royal family members for three generations after Victoria and then disappeared.

Treatment of hemophilia:

Treatment of hemophilia involves regular injections of clotting factor medicine.

Color Blindness

Though most of us share a common color vision sensory experience, some people have a color vision deficiency, which means that their perception of colors is different from what most of us see.

Color-blind people are able to see things as clearly as other people but they are unable to fully ‘see’ red, green, or blue light.

Color blindness is the inability or reduced ability to see colors or to perceive obvious differences between two colors under normal lighting.

Symptoms:

Common symptoms of color blindness include difficulty in distinguishing between colors and inability to see shades or tones of the same color,

1. People with Red-green color blindness are unable to see some shades of red and green,
2. People with Blue-yellow color blindness are unable to see some shades of blue and yellow,
3. People with complete color blindness do not see any colors.

Types:

Color blindness can be mild, moderate, or severe. Based on photo pigments defects in the three different kinds of cone cells that respond to blue, green & red light, color blindness can be of three types

Red-green color blindness:

People with this type of color blindness are unable to see some shades of red and green. Reds may appear brownish-yellow, and greens may look beige to some people.

Certain shades of orange, yellow, and green may appear yellow to others. Red may also appear black to some people. Red-green color blindness is sometimes called Daltonism after John Dalton, the famous scientist, who himself was red-green color blind.

Class 10 Life Science Model Paper WBBSE 2023

Blue-yellow color blindness:

People with this type of color blindness are unable to see some shades of blue and yellow. Blue may appear greener and it may be difficult to distinguish yellow and red from pink. Yellow may appear violet to some people.

Complete color blindness:

People with complete color blindness do not see any colors. Poor vision accompanies complete color blindness.

Red-green color blindness is the most common form of color vision deficiency in many parts of the world. It is also much more common in men than in women.

Blue-yellow color blindness affects men and women equally. Complete color blindness is rare.

Causes:

Color blindness is a genetic condition caused by a difference in how one or more of the light-sensitive cells found in the retina of the eye respond to certain colors.

Inherited forms of color blindness often are related to deficiencies in certain types of cones or outright absence of cones.

Usually, genes inherited from the parents cause the deficiency or are responsible for faulty photopigments.

Men are much more likely to be colorblind than women because the genes responsible for the most common, inherited color blindness are on the X chromosome.

Males only have one X chromosome, while females have two X chromosomes. In females, a functional dominant gene for normal vision on only one of the X chromosomes is enough to compensate for the loss on the other.

In X-linked inheritance, the mother carries the mutated gene on one of her X chromosomes and will pass on the mutated gene to 50 percent of her children.

Inheritance of colour blindness:

In X-linked inheritance, the mother carries the mutated gene on one of her X chromosomes and will pass on the mutated gene to 50 percent of her children.

Because females have two X chromosomes, the effect of a mutation on one X chromosome is offset by the normal gene on the other X chromosome.

In this case, the mother will not have the disease, but she can pass on the mutated gene and so is called a carrier.

If a mother is a carrier of an X-linked dis-ease (and the father is not affected), there is a, 1 in 2 chance that a son will have the disease, a 1 in 2 chance that a daughter will be a carrier of the disease, No chance that a daughter will have the disease.

Treatment:

In general, treatments include blood transfusions, bone marrow transplants, medications, and supplements, etc.

People who receive blood transfusions receive extra iron that the body can’t easily get rid of and iron can accumulate in tissues, which can be potentially fatal.

The human body has no active mechanism for the excretion of iron.

Excess iron in vital organs, even in mild cases of iron overload, increases the risk for liver disease (cirrhosis, cancer), heart attack or heart failure, diabetes mellitus, osteoarthritis, osteoporosis metabolic syndrome, hypothyroidism, etc.

Hence the treatment of blood transfusion is to be followed with Iron chelation which involves the removal of excess iron from the bloodstream.

Inheritance of thalassemia:

A child who inherits two thalassemia trait genes-one from each parent will have the disease.

A child of two carriers has a 25 percent chance of receiving two trait genes and developing the disease and a 50 per cent chance of being a thalassemia trait carrier.

As illustrated in marriages between two carriers (thalassemia minor) may result in – a 25% chance of thalassemia major children, a 50% chance of thalassemia minor children, and a 25% chance of normal children. Thus two thalassemia carriers should not get married.

Question 4. Evaluate the effects of the following pollutants on the environment and human health:

1. Non-biodegradable insecticides
2. Pollen grain
3. Chemical fertilizers containing phosphate and nitrate
4. Wastes containing pathogens originated from health centers
5. Chlorofluorocarbon

Answer:

Effects of Pollutants on the Environment and human health

Or,
Speculate the probable causes of each of the following phenomena:

1. Insomnia, High blood pressure, Partial or Complete deafness
2. Decrease in the concentration of dissolved oxygen in the water and floating of dead fishes
3. Decrease in the number of Crocodiles’ Inflammation in the respiratory tract of the lung
4. Decrease in the number of pollinating insects

Answer:

Probable causes of different phenomena

Question 5. ‘The ever-increasing population in the different cities of India is creating the crisis of ground water’-Support the statement with reasons on the basis of your experiences. Construct a concept map to show how the increase in human habitat is influencing the ecosystem of Sunderbans.

Answer:

Groundwater in aquifers below the surface of the earth is an important natural resource. It accounts for nearly 30% of all fresh water.

Class 10 Life Science Model Paper WBBSE 2023

Depletion of groundwater is caused due to the following reasons:

Overuse:

As the population continues to rise, frequent pumping of groundwater causes depletion since it does not have enough time to replenish itself. This has become a major cause in every big city and town.

Agricultural Needs:

Lack of adequate surface irrigation facilities leads to over-exploitation of groundwater to meet crop and livestock requirements. This is a persistent problem in Gangetic Plain and north India.

Anthropogenic pollution:

Groundwater contamination occurs due to the mixing of gas, oil, poisonous salts, and chemicals or pesticides rendering it unsafe.

The environmental effects of groundwater depletion include land subsidence, water shortage, loss of biodiversity, hampered precipitation rate, food shortage, etc.

(or)

Summarise which conservation measures have been adopted to increase the population of an endangered mammal exclusively found in the swampy grasslands under the foothills of the Eastern Himalayas. Discuss the role of biodiversity in maintaining the balance of the ecosystem of a river.
Answer:

Endangered mammal (Red panda) conservation measures:

Since river or lotic ecosystem refers to flowing water, hence biotic components are specialized to live with flow conditions. They include bacteria, primary producers, insects and other invertebrates, fish, and other vertebrates.

Bacteria decompose the organic matter, vegetation, and biofilm. They play a large role in energy recycling.

Primary producers:

These include phytoplanktons, mosses, liverworts, and many rooted plants. These are sources of energy and form microhabitats that shelter other fauna from predators and the water current.

Insects and other invertebrates act both as consumers and prey items in lotic systems.

Various species of fish and other vertebrates like salamanders, snakes, crocodiles, turtles, bird species, and mammals are part of the lotic ecosystem. They act as consumers and also as prey species to the larger vertebrates available.

WBBSE Chapter 3 Heredity And Common Genetic Diseases Topic A Heredity Introduction To Heredity and Genetics

All living organisms reproduce. Reproduction results in the formation of offspring of the same kind. A pea plant, for example, produces only pea plants each time it reproduces.

Likewise, a rat produces only rats or humans produce only humans. On the other hand, members of a family share many similarities in appearance, such as height, eye color, hair color, etc.

However, the resulting offspring need not and most often do not resemble the parent. Several characteristic differences do occur between individuals belonging to the same species.

It should be noted that the similarities and differences among the members of a species are not coincidental. Both the similarities and differences are received from their parents.

The mechanism of transmission of characters or traits comprising both resemblances as well as differences, from the parental generation to the offspring through reproduction is called heredity.

Read and Learn More WBBSE Solutions For Class 10 Life Science

WBBSE Chapter 3 Topic A Heredity Introduction To Heredity And Genetics Heredity And Variation

The offspring of all the organisms (plants and animals) resemble their parents in several aspects. This is only due to the phenomenon of heredity.

Heredity is the transmission of characters from one generation to successive generations or from parents to their offspring.

Thus, heredity is the cause of similarities between the offspring, so that the individuals of the same parents resemble each other in many aspects.

Heredity involves the transfer of genetic characteristics from parents to the offspring via the egg and sperm.

On the other hand, though offsprings receive all the characteristics of their parents, still they are not exactly alike.

Heredity Class 10 Life Science

Differences are found even between the offspring of the same parents in terms of the shape of faces, hair color, or even skin color.

It is thus difficult to find identical individuals. The progeny differs not only in itself but also with the parents.

These differences are called variations.

Thus, variations may be defined as the differences (morphological, physiological, cytological, and behavioral) between the parents and the offspring or between the offspring of the same parents, family, and race.

Significance Of Variations:

Variations differentiate one individual from another.

Variations enable individuals to adapt themselves according to the changing environment, ie. they make some individuals better fitted or suited to face the struggle for existence.

Discontinuous variations introduce new traits in the species.

Variations are the key to the evolution and development of new species.

WBBSE Chapter 3 Topic A Heredity Types Of Variation

Primarily Variations May Be Classified Into the Following Two Types:

Hereditary Variation:

The variations that arise as a result of any change in the structure and function of the gene and are inherited from one generation to another are called hereditary variations.

Environmental Variations:

Two individuals with the same genotype may become different in phenotype when they come in contact with different conditions of food, temperature, light, humidity, and other external factors.

Such differences among organisms of similar heredity are known as environmental variation. These are not heritable.

Variations may also be classified based on the following parameters:

Based On The Type Of Cells Involved,

Variation Is Classified Into Two Types:

Somatic Variation:

The variation which occurs in somatic cells is called somatic variation. It is generally insignificant because it is not inherited from parents.

It is acquired by the organisms during their lifetime and is lost with death. Hence, it is also called acquired variation.

Somatic variations are caused by three types of factors:

Environmental factors include temperature, light, nutrition, water supply, habitat, topography, enemies, etc.

Use & disuse of organs as may be found in singers whose vocal organs are far more developed than the non-singers.

Conscious efforts include receiving education, developing certain good or bad, or unhealthy habits, etc.

Germinal Variation:

The variation which affects the germinal or reproductive cells is called germinal variation. It is heritable and genetically significant.

It provides raw materials for evolution.

Heredity Class 10 Life Science

The factors that cause germinal variations include chance separation of chromosomes, a chance combination of chromosomes, crossing over, chromosomal aberrations, change in chromosome number, genetic mutations, etc.

Based On The Degree Of Differences, Variation Is Classified Into Two Types:

Continuous Variation:

Small and indistinct variations are called continuous variations, eg. the shape of the nose or the color of the skin, etc.

These fluctuate with environmental conditions.

These are non-heritable.

They have no role in evolution.

They are most common and occur in all organisms or races of a species.

Discontinuous Variation:

Large, distinct, and sudden variations are called discontinuous variations, or mutations, eg. the appearance of six fingers in a man, polydactyly, etc.

These are relatively unaffected by environmental conditions.

These are mostly heritable. However, not all the discontinuous variations pass to the next generations because of their appearance in the body cells after the differentiation of germ cells.

They provide raw materials for evolution on which selection is based.

They are not common and appear suddenly.

Based on the affected traits, variations may be of four types:

Morphological variation:

These are the differences that are found in the form and structure of organisms.

Physiological variation:

These are the variations observed in various functions of the body like BMR etc.

Cytological variation:

These variations occur in the number of cells, cell constituents, and their products.

Behaviouristic variation:

These variations are connected with the behavior of the organisms towards different conditions.

Based on impact, variations are of three types:

Beneficial Variation:

These are the variations that help the organisms skillfully adapt to the particular environment.

Heredity Class 10 Life Science

Harmful variations:

These variations make organisms unfit for their environment.

Neutral variations:

These variations do not affect the organisms in any way.

WBBSE Chapter 3 Topic A Heredity Sources Of Variation

Variations are more pronounced in sexually reproducing organisms. In sexual reproduction, the production of offspring takes place by the fusion of two types of gametes.

These gametes are formed by the reduction division. Thus, sexual reproduction introduces unlimited genetic variation into the population. The more closely related the sexual partners are, the lesser the variations in their offspring.

Common sources of variations are-

Recombination:

Recombination results in offspring that have a combination of characteristics different from that of their parents. Different types of combinations of characters bring about variation.

Crossing Over:

At the time of gamete formation, crossing over occurs during meiosis division, which causes variations in genetic characteristics.

Mutation:

It is a spontaneous, sudden, heritable, and permanent change in genetic characteristics that causes a detectable effect in the organisms. It occurs in nature or it can be caused artificially in an organism.

WBBSE Chapter 3 Topic A Heredity Mendel The Father Of Genetics

There have been several explanations on the possible mechanism of inheritance of traits from the parent to the offspring, put forth from time to time by different biologists.

All these early theories presume that the characteristics of the two parents somehow mix during inheritance. Hence these ideas came to be known as blending theories of inheritance.

Gregor Johann Mendel (1822-1884) was born in a family of poor peasants in Moravia, Austria. He received his school education with utmost difficulty due to poverty in the family.

Wbbse Class 10 Life Science Heredity Notes

In 1843 he joined a church as a monk where, in 1847, he became the abbot (head) of the monastery at Brunn, Austria (now called Bruno in Czechoslovakia).

In addition to his normal duties of preaching in the church, Mendel evinced a keen interest in the maintenance of the garden in the premises of the church.

In the course of his routine rounds in the garden, Mendel was keenly observing the pattern of inheritance of certain characters in some of the plants.

He became interested in investigating the mechanism by which the characters are transferred from the parent plants to their offspring. He decided to conduct some experiments in this direction.

After careful examination and thinking, Mendel selected the pea plants (Pisum sativum) for his experiments.

Gregor Mendel, through his work on pea plants, discovered the fundamental laws of inheritance.

He deduced that genes come in pairs and are inherited as distinct units, one from each parent. Mendel tracked the segregation of parental genes and their appearance in the offspring as dominant or recessive traits.

He recognized the mathematical patterns of inheritance from one generation to the next.

Thus, Mendel’s concept gave birth to the particulate theory of inheritance. Mendel’s excellent experiments, valid mathematical analysis, and formulation of laws of inheritance are collectively known as Mendelism.

The genetic experiments Mendel did with pea plants took him eight years (1856-1863) and he published his results in 1865. During this time, Mendel grew over 10,000 pea plants, keeping track of progeny number and type.

Wbbse Class 10 Life Science Heredity Notes

Mendel’s work and his Laws of Inheritance were not appreciated in his time. As a result, Mendel died in 1884 without any sort of recognition.

It was not until 1900, after the rediscovery of his Laws, that his experimental results were understood.

Three botanists – Hugo DeVries, Carl Corrensand Erich von Tschermak – independently rediscovered Mendel’s work in the year 1900, a generation after Mendel published his papers.

They helped expand awareness of the Mendelian laws of inheritance in the scientific world. The three Europeans, unknown to each other, were working on different plant hybrids when they each worked out the laws of inheritance.

WBBSE Chapter 3 Topic A Heredity Hereditary Variation

The variations that arise as a result of any change in the structure and function of the gene and that are inherited from one generation to another are called hereditary variations.

The ultimate source of all genetic variation is mutation. It leads to changes in gene function and permanent alteration to the DNA sequence.

It is a rare, random change in the genetic material and it can be inherited. The permanent alteration in the DNA sequence that makes up a gene results in difference which is found in most people.

Mutations range in size; they can affect anywhere from a single DNA building block (base pair) to a large segment of a chromosome that includes multiple genes.

This is the only way new alleles (varieties of a gene) are produced. Mutations are rare events: the average rate of mutation is about one per 100,000 genes.

Wbbse Class 10 Life Science Heredity Notes

Thus it would take on average 100,000 generations for a mutation to occur at any one specific gene. However, each individual has many, many functional genes.

Thus, at the level of the whole individual, mutations occur quite often. It is estimated that each human gamete (egg or sperm cell) has, on average, one mutation.

The heritable change in the composition of a gene that leads to the formation of a mutant gene having changed function and with the consequent appearance of a new phenotype is known as mutation.

Types Of Mutations:

There Are Two Types Of Mutations:

Gene mutations or point mutations, and

Chromosomal mutations.

1. Gene mutations:

A chemical change that occurs in the DNA of a cell is called a gene mutation or point mutation. Such a mutation may alter the sequence of the nucleotides within a part of the DNA molecule.

This alternation changes the information of the DNA chain and results in differences in the proteins being produced.

For Example:

In sickle cell anemia, the mutation of a single gene causes a slight change in the structure of the protein molecule of hemoglobin, and because of that slight change, the blood cell that carries the hemoglobin takes a sickle shape.

According to many scientists, the mutations may be caused naturally by the radiation that constantly enters the earth’s atmosphere from the cosmos.

For example, gene mutations are probably caused when the sex cells (gametes) of an organism are exposed to X-rays, gamma rays, cosmic rays, and ultraviolet rays. In addition to this radiation, certain chemicals called mutagens can change nucleotides within DNA molecules.

Wbbse Class 10 Life Science Heredity Notes

The chance for a particular human gene to mutate in one generation is about 1 in 10,000 to 1 in 1,000,000. Since humans have at least 30,000 genes, each person likely carries at least one mutation.

2. Chromosomal Aberrations (Chromosome mutations):

Another way for the genetic traits of an organism to be altered is through changes involving whole chromosomes or parts of chromosomes.

Structural changes in chromosomes are also caused by radiation, chemicals, and even some virus infections.

Chromosomal mutations may be of two basic types-

1. Change of structure, and
2. Change of number.

Chromosome structure changes:

Changes occur in the structure of chromosomes, during cell division. When homologous chromosomes pair up, linked genes on the chromosomes may break apart. The genes may join another chromosome, or they may be lost.

Thus, a deletion involves the loss of a piece of chromosome.

If a chromosome breaks and the parts do not reattach, the pieces may be lost. This is the most serious kind of chromosome mutation. Here, bits of genetic information are not available to the offspring.

Duplication occurs when one extra, but identical piece of a chromosome is added to the normal chromosome When an inversion occurs, the pieces of chromosomes break apart and pieces rejoin the same chromosome in a different order. Usually, inversions have no harmful effects on the offspring.

The joining of a fragmented chromosome to a non-homologous chromosome is a translocation. The piece of chromosome detaches from one chromosome and moves to a new position on another chromosome.

Chromosome number changes:

A chromosome mutation that causes individuals to have an abnormal number of chromosomes is termed aneuploidy. Aneuploid cells occur as a result of chromosome breakage or non-disjunction errors that happen during meiosis or mitosis.

Non-disjunction is the failure of homologous chromosomes to separate properly during cell division. It produces individuals with either extra or missing chromosomes.

Sex chromosome abnormalities that result from non-disjunction can lead to conditions such as Klinefelter and Turner syndromes. In Klinefelter syndrome, males have one or more extra X chromosomes.

In Turner syndrome, females have only one X chromosome. Down syndrome is an example of a condition that occurs due to non-disjunction in autosomal (non-sex) cells. Individuals with Down syndrome have an extra chromosome on autosomal chromosome 21.

Mendel’S Laws Of Heredity Class 10

A chromosome mutation that results in individuals with more than one haploid set of chromosomes in a cell is termed polyploidy. A haploid cell is a cell that contains one complete set of chromosomes.

Our sex cells are considered haploid and contain 1 complete set of 23 chromosomes. Our autosomal cells are diploid and contain 2 complete sets of 23 chromosomes.

If a mutation causes a cell to have three haploid sets, it is called triploidy. If the cell has four haploid sets, it is called tetraploidy.

Due to the mutagens, two types of mutations are found visible and lethal mutations. Mutations are located on either sex chromosomes or autosomes.

Common Variations In Human Population

Some examples of often un-noticed human traits are the ability or inability to roll the tongue, attached or unattached earlobes, dimples or freckles, naturally curly or straight hair, smooth or cleft chin, color blindness or normal vision, etc.

There are numerous traits in humans, but some traits occur more frequently than others. Between 70-90% of the human population have free-hanging earlobes, can roll their tongue, are right-handed, and can taste a chemical called PTC (Phenyl-thio-carbamide).

But these characters have nothing to do with life efficiencies.

Some common examples are-

Ear Lobe:

Some people have earlobes that curve up between the lowest point of the earlobe and the point where the ear joins the head; these are known as “free” or “unattached” earlobes.

Other people have earlobes that blend in with the side of the head, known as “attached” or “adherent” earlobes [Fig 3.3(b)]. Attached vs.

free earlobes are often used to illustrate basic genetics. The major form of the gene that determines the shape of the earlobe is known as an allele. An allele is a gene that is found at a specific position on a chromosome.

If the genes from the parents get expressed by the dominant allele, then the child will be born with free earlobes.

The structural formation of the attached earlobe is due to the absence of the dominant allele in the chromosomes. The recessive allele is expressed in the chromosomes to form an attached earlobe.

Thus, if a person is homozygous recessive for this trait, the earlobes attach directly to the head and do not hang free. However, it is not necessary that parents with attached earlobes should give birth only to the attached earlobe child.

If, on the other hand, parents with free earlobes give birth to a baby with attached earlobes, it is certain that both of them have a copy of the dominant and recessive alleles.

Tongue Rolling:

Tongue rolling is the ability to roll the lateral edges of the tongue upwards into a tube. A dominant allele enables some people to roll their tongues into a distinct U-shape. If you cannot roll your tongue you carry the recessive trait.

Recent studies have shown that this tongue-rolling could be dependent on multiple genes or alleles & some kind of environmental influences.

WBBSE Chapter 3 Topic A Heredity Key Terms Associated With Heredity

Some basic key terms associated with heredity and genetics are—

Characteristics Or Trait:

A characteristic or phenotypic trait, or simply trait, is a distinct variant of a phenotypic characteristic of an organism that may be inherited, environmentally determined, or maybe a combination of the two.

For example, eye color is a character or abstraction of an attribute, while blue, brown, and hazel are traits.

Individuals and groups differ among themselves biologically, in a practically endless succession of more or less visible elements of their descriptions, which are named as traits, features, marks, nature, characteristics, characters, and others.

Mendel’s Laws Of Heredity Class 10

Each of these components is a description of, say, the observational nature, i.e. selected part of our vision or measuring the actual condition of the individual body or group structure.

Allele:

An allele is an alternative or some form of a gene (one member of a pair) that is located at a specific position (locus) on a specific chromosome.

These DNA codings determine distinct traits that can be passed on from parents to offspring through sexual reproduction. Alleles may occur in pairs, or there may be multiple alleles affecting the expression (phenotype) of a particular trait.

The word “allele” is a short form of allelomorph (“other form”), which was used in the early days of genetics to describe variant forms of a gene detected as different phenotypes.

Organisms generally have two alleles for each trait, for example- yellow or green seed colors of pea plants; in humans, the gene for eye color has an allele for blue eye color and an allele for brown.

For any gene, a person may have the same two alleles or two different ones.

Locus:

In genetics, a locus (plural loci) is the specific location of a gene, DNA sequence, or position on a chromosome.

Each chromosome carries many genes; in humans, the estimated number of ‘haploid’ protein-coding genes is 20,000-25,000, located on 23 different chromosomes. A variant of a similar DNA sequence located at a given locus is called an allele.

A locus is the specific physical location of a gene or other DNA sequence on a chromosome, like a genetic street address.

A locus is a spot or “address” on a chromosome at which a gene for a particular trait is located in all members of a species. It can also refer to the location of a mutation or other genetic marker.

A given locus can be found on any pair of homologous chromosomes. For example, in Drosophila, the locus of red and white eye colors is present on the X chromosome.

Unit of inheritance (Factor/Gene):

A gene is a stretch of DNA that helps to control the development and function of all organs and working systems in the body.

Mendel didn’t know about genes or discover genes, but he did speculate that there were two factors for each basic trait and that 1 factor was inherited from each parent.

We now know that Mendel’s inheritance factors are genes (the term was first used by Johannsen, 1909), or more specifically alleles – different variants of the same gene. It is now known that Mendelian factors, determinants, or genes are present in a linear sequence on the chromosomes.

Therefore, the gene is also defined as a unit of inheritance that consists of linear chromosome segments that can be assigned to the expression of a particular Character.

Mendel’s Laws Of Heredity Class 10

Its effect is, however, influenced by its allele, other genes, and the environment.

Genes are passed from parent to offspring; the combination of these genes affects all aspects of the human body, from eye and hair color to how well the liver can process toxins.

Monohybrid and dihybrid cross:

A monohybrid cross is a breeding experiment between P-generation (parental generation) organisms that differ in one trait.

It is a genetic cross between parents that differ in the alleles they possess for one particular gene, one parent having two dominant alleles and the other two recessives.

All the offspring (called monohybrids) have one dominant and one recessive allele for that gene (i.e. they are hybrid at that one locus). Generally, the monohybrid cross is used to determine the dominance relationship between two alleles.

Example:

The allele for green pod color is dominant and the allele for yellow pod color is recessive. The cross-pollination between a P-generation green pod pea plant and a P-generation yellow pod plant results in all green offspring.

Crossing between these offspring yields a characteristic 3:1 (monohybrid) ratio in the following generation of dominant recessive phenotypes.

A dihybrid cross is a cross between two pure lines (varieties, strains) that differ in two observed traits. A dihybrid cross describes a mating experiment between two organisms that are identically hybrid for two traits.

A hybrid organism is heterozygous, which means that it carries two different alleles at a particular genetic position or locus.

Therefore, a dihybrid organism is heterozygous at two different genetic loci. Mendel used these results as the basis for his Law of Independent Assortment.

Example:

Mendel performed a dihybrid cross using pea plants and the characteristics of seed color and texture: the parental plants had either smooth yellow seeds- the dominant characteristics – or wrinkled green seeds- the recessive characteristics.

All the offspring had smooth yellow seeds, being heterozygous for the two alleles. Crossing between these offsprings produced an F₂ generation of plants with smooth yellow, smooth green, wrinkled yellow, and wrinkled green seeds in the ratio of 9:3:3:1.

Homozygous and heterozygous organisms:

Organisms can be homozygous or heterozygous for a gene. Homozygous means that the organism has two copies of the same allele for a gene.

An organism can be homozygous dominant if it carries two copies of the same dominant allele, or homozygous recessive if it carries two copies of the same recessive allele.

Heterozygous means that an organism has two different alleles of a gene. For example, pea plants can have red flowers and either be homozygous dominant (red-red), or heterozygous (red-white).

If they have white flowers, then they are homozygous recessive (white-white). Carriers are always heterozygous.

Example:

An organism is referred to as being homozygote or homozygous at a specific locus when it carries two identical copies of the gene affecting a given trait on the two corresponding homologous chromosomes (e.g., the genotype is PP or pp when P and prefers to different possible alleles of the same gene).

Such a cell or such an organism is called a homozygote.

An organism is a heterozygote or is heterozygous at a locus or gene when it has different alleles occupying the gene’s position in each of the homologous chromosomes.

In diploid organisms, the two different alleles are inherited from the organism’s two parents. For example, a heterozygous individual would have the allele combination Pp.

Hybridization:

Genetic hybridization is the process of interbreeding individuals from genetically distinct populations to produce a hybrid. A genetic hybrid would therefore carry two different alleles of the same gene.

During the 20th century planned hybridization between carefully selected parents has become dominant in the breeding of self-pollinated species.

The object of hybridization is to combine desirable genes found in two or more different varieties and to produce pure-breeding progeny superior in many respects to the parental types.

The process of hybridization is important biologically because it increases the genetic variety (number of different gene combinations) within a species, which is necessary for evolution to occur.

Mendel’s Laws Of Heredity Class 10

Example: Hybridization is the process of crossing two genetically different individuals to create new genotypes.

For example, a cross between parent 1, with the genetic makeup (genotype) BB, and parent 2, with bb, produces progeny with the genetic makeup Bb, which is a hybrid (the first filial generation or FI).

Hybridization was the basis of Gregor Mendel’s historic experiments with garden peas. Inheritance studies require crossing plants with contrasting or complementary traits.

Pure and hybrid:

A diploid organism has paired chromosomes, each with a similar arrangement of genetic loci.

Variations of these genes are called alleles. If an organism has one of the same types of alleles on each of its chromosomes, that organism has a pure trait.

If an organism has two different types of alleles on its chromosomes, that organism has a hybrid trait. Pure breed or true breeding are individuals that are homozygous and that will always produce the same offspring when crossed together.

A hybrid is an organism that has two different alleles for a trait.

In the simplest possible terms, purebreds are the offsprings that result from mating between genetically similar parents while hybrids are the offsprings that are the result of mating between two genetically dissimilar parents.

Thus, purebreds are composed of two (or more) like components while hybrids are created using two or more similar but not like components.

Parental generation:

The parental generation (P) is the first set of parents crossed. It is the generation of individuals of different genotypes that are mated, usually for scientific purposes, to produce hybrids.

These parental strains are purified and obtained by repeated cycles of self-fertilization. In a parental generation, two individuals are mated or crossed to determine or predict the genotypes of their offspring, called the first filial generation.

Immediate parents are designated as Pv grandparents P2; great grandparents are P3 and so on.

Filial generation:

Filial generation is the offspring generation. It is a generation in a breeding experiment that is successive to mating between parents of two distinctively different but usually relatively pure genotypes.

F₁ is the first offspring or filial generation; F₂ is the second; and so on.

Successive generations of progeny result in a controlled series of crosses, starting with two specific parents (the P generation) and selfing or intercrossing the progeny of each new (F₁F₂; etc. ) generation.

Filial 1 (F₁) generation is the one resulting from the cross of such two selected parent generations; this generation expresses a high degree of uniformity (hybrids).

The progeny/generation derived from the Filial 1 generation is termed the Filial 2 generation (F₂); this generation can show a very high degree of variation from progeny to progeny depending upon the parental generation used.

Dominant and recessive characteristics:

The terms dominant and recessive describe the inheritance patterns of certain traits. Sexually reproducing species, including people and other animals, have two copies of each gene.

The two copies, called alleles, can be slightly different from each other. The differences can cause variations in the proteins that are produced.

Proteins affect traits, so variations in protein activity or expression can produce different phenotypes. A dominant allele produces a dominant phenotype in individuals who have one copy of the allele, which can come from just one parent.

For a recessive allele to produce a recessive phenotype, the individual must have two copies, one from each parent.

An individual with one dominant and one recessive allele for a gene (heterozygous) will have the dominant phenotype.

Heredity Class 10 Life Science

For example, in humans, the allele for brown eyes is dominant, therefore offspring only need one copy of the ‘brown eye’ allele to have brown eyes (although, with two copies they will still have brown eyes).

Recessive alleles only show their effect if the individual has two copies of the allele (also known as being homozygous).

For example, the allele for blue eyes is recessive, therefore to have blue eyes the offsprings need to have two copies of the ‘blue eye’ allele.

Phenotype And Genotype:

Genotype is the genetic makeup of an individual organism that functions as a set of instructions for the growth and development of the body.

The word ‘genotype’ is usually used when talking about the genetics of a particular trait (like eye color).

Phenotype is the observable physical or biochemical characteristics of an individual organism, determined by both genetic makeup and environmental influences, for example, height, weight, and skin color.

An organism’s genotype is the set of genes that it carries. An organism’s phenotype is all of its observable characteristics—which are influenced both by its genotype and by the environment.

The “internally coded, inheritable information”, or Genotype, carried by all living organisms, holds the critical instructions that are used and interpreted by the cellular machinery of the cells to produce the “outward, physical manifestation”, or Phenotype of the organism.

The entire set of genes or the genotype in a black mouse. It comprises a set of alleles that determines the expression of a particular characteristic or trait (phenotype). Her black hair is the trait or phenotype concerned.

All the key terms described in this article (3A.2), are explained with sample crosses in the following articles.

WBBSE Chapter 3 Topic A Heredity Mendel’s Work On Pea Plants

In 1854, Mendel began a series of breeding experiments with the garden pea Pisum sativum in an attempt to learn something about the mechanisms of heredity.

The bisexual flower of pea normally reproduces by self-fertilization.

That is, the stamen (male reproductive organs) produces pollens, which land on the pistil (female reproductive organs) within the same flower and consequently fertilize the plant. This process is called selfing.

To prevent selfing and to carry out cross-pollination in pea plants, three steps of emasculation, dusting and bagging are to be carried out as elaborated here:

It is relatively a simple procedure to prevent self-fertilization of the pea by removing the stamens from a developing flower bud before they produce any mature pollen.

For this purpose, the stamens are cut off by a scissor from the selected bisexual flowers to remove the masculine part of the flower.

This process is known as emasculation. The pollen taken from the stamens of another flower can then be dusted onto the stigma of the pistil of the emasculated one to cross-pollinate it. After cross-pollination, all the

flowers are covered with paper bags to avoid any contamination by undesired pollen grains.

Cross-fertilization, or more simply a cross, is the term used for the fusion of male gametes (pollen) from one individual and female gametes (eggs) from another.

Once cross-fertilization has occurred, the zygote develops in the seeds (peas), which are then planted.

For his experiment, Mendel obtained 34 strains of pea plants that differed in a number of traits.

He allowed each strain to self-fertilize for many generations to ensure that he only worked with pea strains in which the trait under investigation remained unchanged from parent to offspring for many generations.

Wbbse Class 10 Life Science Heredity Notes

Such strains are called true-breeding or pure-breeding strains.

In Mendel’s work true breeding or pure breeding homozygous strains constitute the parental (P) generation.

Once cross-fertilization is carried out in a parental generation, the progeny that is obtained is known as the First generation offspring or First filial (F₁) generation.

WBBSE Chapter 3 Topic A Heredity Mendel’s Experiment And Laws For Monohybrid Cross

Mendel performed several mono-hybridization experiments on pea plants involving crosses between parents that differ in a single gene.

WBBSE Chapter 3 Topic A Heredity The seven pairs of characters as chosen by Mendel

Mating between individuals that differ in only one trait, such as seed color or stem length is known as a monohybrid cross.

In each monohybrid cross, one parent carries one form of the trait, and the other parent carries an alternative form of the same trait. Mendel selected seven such traits to study the monohybrid breeding experiment.

Each trait had two easily distinguishable, alternative appearances (phenotypes).

These are—

WBBSE Chapter 3 Topic A Heredity Experiment And Observation

Mendel carried out a series of monohybrid crosses.

For example, in the spring of 1854, he planted pure-breeding green peas and pure-breeding yellow peas and allowed them to grow into the parental (P) generation.

Later that spring when the plants had flowered, he dusted the female stigma of green-pea plant flowers with pollens from yellow-pea plants. He also performed the reciprocal cross between the female yellow pea plant and the male green pea plant.

In the fall (autumn), when he collected and separately analyzed the progeny peas he found that in both cases, the peas were all yellow.

The yellow peas, the progeny of the P generation, were the beginning of what we now call the first filial (Fx) generation. Mendel planted them and allowed the F₁ plants to self-fertilize.

He then harvested and counted the peas of the resulting second filial (F₂) generation, the progeny of the Fx generation.

Among the progeny of one series of F₁ self-fertilization, there were 6022 yellow and 2001 green peas, an almost perfect ratio of 3 yellow to 1 green.

The results of reciprocal crosses produced a similar ratio. Monohybrid crosses involving other traits (such as long and short stem length) also showed similar results.

WBBSE Chapter 3 Topic A Heredity Mendel’s explanation

To explain these results, Mendel proposed the existence of what he called particulate unit factors for each trait. He suggested that these factors served as the basic unit of heredity and passed unchanged from generation to generation, determining various traits expressed by each individual plant.

WBBSE Chapter 3 Topic A Heredity Mendel’s First Three Postulates

Using the constant pattern of result in the monohybrid crosses, Mendel derived the following three postulates or principles of inheritance

Unit Factors in Pairs:

Genetic characters are controlled by unit factors existing in pairs in individual organisms.

In the monohybrid cross involving yellow (Y) and green (y) seeds, a specific unit factor exists for each trait. Each diploid individual receives one factor from each parent.

Because the factors occur in pairs, three combinations are possible:

YY, Yy, and Yy. The seeds having unit factor combinations of either YY or yy have two copies of the same allele for a gene. Hence these are homozygous.

The seeds having a unit factor combination of Yy have two different alleles of a gene. Hence these are heterozygous.

Dominance/Recessiveness:

When two unlike unit factors, responsible for a single character, are present in a single individual, one unit factor is dominant to the other, which is said to be recessive.

In each monohybrid cross, the trait expressed in the F₁ generation results from the presence of the dominant unit factor.

The trait that is not expressed in the F₁ but which reappears in the F₂ is under the genetic influence of the recessive unit factor.

In the above-mentioned case, the trait yellow seed color (Y) is said to be dominant to the recessive trait, green seed (y). Hence, the progeny of P, i.e. the F₁ generation had all yellow pea plants.

Among the progeny of ie. in the F₂ generation, green pea plants reappeared along with yellow pea plants.

Segregation:

During the formation of gametes, the paired unit factors responsible for contrasting traits do not blend with each other but separate or segregate randomly, so that each garnet receives one or the other with equal likelihood.

This is known as the 1st law of Mendel or the law of Segregation.

WBBSE Chapter 3 Topic A Heredity Analysis Of Monohybrid Cross And Checker Board

WBBSE Chapter 3 Topic A Heredity Punnett Squares

The genotypes and phenotypes resulting from the recombination of gametes during fertilization can be easily visualized by constructing a Punnett square, named after R. C.

Punnett who first devised this approach. In this construction, each of the possible gametes is assigned an individual column or a row, with the vertical column representing those of the female parent and the horizontal row those of the male parent.

After the gametes are entered in rows and columns, the new generation can be predicted by combining the male and female gametic information for each combination and entering the resulting genotype in the boxes. This process represents all possible random fertilization events.

WBBSE Chapter 3 Topic A Heredity Mendel’s Experiment And Laws For Dihybrid Cross

As a natural extension of the monohybrid cross, Mendel also designed experiments in which he examined two characters simultaneously. Such a cross, involving two pairs of contrasting traits, is called a dihybrid cross.

WBBSE Chapter 3 Topic A Heredity Experiment and Observation

Mendel-crossed pea plants that are heterozygous for two genes at the same time.

To construct such a dihybrid, he mated true-breeding plants grown from yellow round peas (YYRR) with true-breeding plants grown from green wrinkled peas (year).

From this cross, he obtained a dihybrid F₁ generation (YyRr) showing only the two dominant phenotypes, yellow and round. He then allowed these F₁ dihybrids to self-fertilize to produce the F₂ generation.

When Mendel counted the F₂ generation of one experiment, he found 315 yellow round, 101 yellow wrinkled, 108 round green, and 32 wrinkled green peas. There were, in fact, yellow wrinkled and green round recombinant phenotypes, providing evidence that some shuffling of alleles had taken place.

Explanation

From the observed ratios, Mendel inferred the biological mechanism of shuffling the independent assortment of gene pairs during gamete formation.

Because the genes for peas’ color and for shape assort independently, Y can be with R or r in any gamete with equal probability.

Thus, the presence of a particular allele of one gene, say, the dominant Y for pea color, provides no information whatsoever about the alleles of the second gene.

That is, the allele for pea shape in Y carrying game could with equal likelihood be either R or r.

Each dihybrid of the F₁ generation can, therefore, make four kinds of gametes:

YR, Yr, yR, and yr. In a large number of gametes, the four kinds will appear in an almost perfect ratio of 1:1:1:1.

At fertilization then, in a mating of dihybrids, 4 different kinds of eggs can combine with any one of 4 different kinds of pollen, producing a total of 16 possible zygotes in the F₂ generation.

Wbbse Class 10 Life Science Heredity Notes

Once again, a Punnett square is a convenient way to visualize the process.

In fact, there are only nine different F₂ genotypes — YYRR, YYRr, YyRr, YyRR, yyRR, yyRr, YYrr, Yyrr, and yyrr — because the source of the alleles (egg or pollen) does not make any difference.

If we look at the combination of the traits determined by nine genotypes, we will see only four phenotypes— yellow round, yellow wrinkled, green round, and green wrinkled — observed in a ratio of 9 : 3 : 3: 1.

If, however, we look at just pea color or just pea shape, we can see that each trait is inherited in the 3: 1 ratio as predicted by Mendel’s law of segregation.

Punnet Square or Checker Board of Dihybrid cross

WBBSE Chapter 3 Topic A Heredity Mendel’s Fourth Postulate

The above analysis became the basis of Mendel’s second general principle or the fourth postulate, the law of independent assortment.

Independent Assortment:

During gamete formation, segregating pairs of unit factors assort independently of each other and undergo random recombination in all possible combinations governed by chance alone.

This postulate stipulates that any pair of unit factors segregate independently of all other unit factors. Thus, according to the postulate of independent assortment, all possible combinations of gametes will be formed in equal frequency.

WBBSE Chapter 3 Topic A Heredity Reasons for Mendel’s Success

Mendel’s success was dependent upon the following factors:

First, he chose the garden pea (Pisum sativum) as his experimental organism.

These plants can easily be cultivated, crossed, and for each successive generation, Mendel could thus obtain large members of individuals within a relatively short growing season.

By comparison, if he had worked with sheep, each mating would have generated only a few offspring and the time between generations would have been several years.

Second, Mendel examined the inheritance of clearcut contrasting forms of particular traits — round versus wrinkled seed, yellow versus green pod color, etc.

Using such ‘either-or’ traits, he could distinguish and trace unambiguously the transmission of one or the other observed characteristics, because there were neither any intermediate forms nor any of these characters located on separate chromosomes.

Third, Mendel isolated and perpetuated lines of peas that breed true. Mating with such pure breeding lines produce offsprings carrying specific parental traits that remain constant from generation to generation.

Fourth, Mendel carefully controlled his matings, going to great lengths to ensure that the progeny he observed really resulted from the specific fertilization he intended.

Thus he painstakingly prevented the intrusion of any foreign pollen and assured self or cross-pollination as the experiment demanded.

He also performed reciprocal crosses, in which by reversing the traits of male and female parents, he efficiently controlled the path of transmission of a particular trait either via the egg cell within the ovule or via the pollen as per experimental demand.

Fifthly, Mendel worked with a large number of plants, counted and subjected his findings to statistical analysis, and then compared his results with predictions based on mathematical models.

Wbbse Class 10 Life Science Heredity Notes

Finally, Mendel was a brilliant practical experimentalist. He could call and observe an optimum number of individuals from the limited space of the monastery garden.

In short, Mendel purposely set up a simplified ‘black and white’ experimental system and then successfully out how it worked.

Genetic Crosses With Guineapig

Mendel worked on pea plants but the application of his laws on animals was carried out by his successors.

WBBSE Chapter 3 Topic A Heredity Monohybrid Cross

When a pure (homozygous) black-haired guineapig (BB) is crossed with a pure white-haired guineapig (bb), all the F₁ offsprings (Bb) are found to be black-haired despite the presence of two contrasting genes for black hair and white hair.

It means black color is dominant to white color in guineapig and F₁ black offsprings (Bb) are heterozygous. Here the gene for the black color is represented by B and for the recessive white color by b.

When the F₁ heterozygous guineapigs are bred among themselves, individuals of the F₂ generation are produced. The F₂ phenotypic ratio is 3 (black): 1 (white) whereas the F₂ genotypic ratio is (pure black, BB) 1: (heterozygous black, Bb) 2: (pure white, bb) 1.

This shows that the inheritance of hair color in guinea pigs follows Mendel’s first law of Segregation.
The monohybrid cross in guinea pig is shown here with a checkerboard.

WBBSE Chapter 3 Topic A Heredity Dihybrid Cross

In guinea pigs, the black coat (BB) is dominant to the white coat (bb), and short hair (SS) is dominant over long hair (ss).

When a homozygous black short-haired male guinea pig (BBSS) is crossed with a white long-haired female (bass), all the F₁ offspring obtained are with black short hair (BbSs).

This shows the dominance of black coats over white coats and the dominance of short hair over long hair.

When the hybrid individuals are allowed to interbreed, F₂ generation consisting of four types of individuals is produced in the following phenotypic ratio = 9 (black short) : 3 (black long) : 3 (white short): 1 (white long).

The F₁ hybrids (BbSs) produced gametes when Bb and Ss alleles were segregated and assorted independently to produce four types of gametes: BS, Bs, bS, and bs.

Thus four types of male gametes and similar four types of female gametes are produced.

These gametes undergo fertilization at random to produce 16 different types of zygotes in the F₂ generation having the above-mentioned four types of phenotypes. This clearly explains Mendel’s law of independent assortment.

WBBSE Chapter 3 Topic A Heredity Deviation Of Menders Laws Of Heredity

Although Mendel formulated the postulates that provide the basis of our understanding of genetic principles, there are many types or modes of inheritance that Mendel simply didn’t encounter.

These modes of inheritance were encountered when investigators began using Mendel’s postulates to study inheritance in other organisms.

Some of these modes of inheritance appear, at first glance, to obey different rules than those that Mendel proposed. So, was Mendel wrong then? Not really.

Wbbse Class 10 Life Science Heredity Notes

His postulates fit the data that he collected, but as stated above, there are situations that he didn’t observe in his study of the pea.

This is a common occurrence in science: when a theory is proposed, it is used as long as it is useful in explaining some aspect of nature.

When data are encountered that don’t fit the theory, the theory may have to be modified (if possible) or discarded. Mendel’s ideas as originally presented may not fit every possible mode of inheritance, but they still provide the basis for explaining those other types of inheritance.

WBBSE Chapter 3 Topic A Heredity Genetic Interaction

According to Mendel each trait or character is controlled by a pair of factors or genes. But later discoveries proved that in many cases the expression of a single character is controlled by the interaction of more than one pair of genes.

The coordinated effect of two or more genes in producing a given phenotypic trait is known as genetic interaction. It was proposed by Bateson and Punnet in the form of a factor hypothesis.

This hypothesis states that certain characters are controlled by the interaction of two or more genes.

The interaction of genes may be classified into two types:

1. Non-allelic and
2. Allelic gene interaction.

The genetic interactions that occur between genes located in the same chromosome or different chromosomes are known as nonallelic gene interactions.

For instance, in some animals, a gene at one locus on a chromosome totally suppresses the expression of a gene at another locus. Such genes are called inhibiting genes and the phenomenon is known as epistasis, eg. ABO blood group in man.

There may be supplementary genes that interact in such a way that one dominant gene produces its effect whether another dominant gene is present or not, but when the second dominant gene is added to the first, a new character is expressed; eg. coat color in mice.

There may also be collaborator genes that influence the same trait but interact to produce totally new traits that neither of the genes could produce, eg. comb shape in fowls.

The other type of genetic interaction that occurs between the two alleles of a single type of gene is known as allelic gene interaction.

For instance, if genes are within a certain distance on the same chromosomes, they do not follow the Law of independent assortment; instead, they are linked when transmitted to the next generation.

This is called linkage. Genes that are located in the cytoplasm do not follow either of Mendel’s Laws, they exhibit maternal inheritance traits.

If the dominance is incomplete, a dominant trait wouldn’t be observed immediately. Sometimes genes become co-dominant, meaning both alleles will show a phenotype.

There may be multiple genes in which case two or more independent pairs of factors affect the same characters but in an additive manner, eg. human skin color.

WBBSE Chapter 3 Topic A Heredity Incomplete Dominance

A common example of deviation from Mendelism is the phenomenon called incomplete dominance.

A cross between parents with contrasting traits may sometimes generate offspring with an intermediate phenotype.

In a heterozygote organism carrying both a dominant and a recessive allele of the same gene, when the dominant gene cannot express its dominant phenotype completely, a mixed or intermediate, or blended phenotype between the dominant and the recessive is expressed.

Such a situation is known as incomplete dominance. In many plant species, flower color serves as a striking example of incomplete dominance.

With the flowers of Four O’Clocks or floret clusters of Snapdragons Mirabilis jalapeno, a cross between pure breeding red flowered parents and pure breeding white yields hybrids with pink blossoms.

During gametogenesis, the pure red flowered parent plant (AA) produces (A) gametes and the pure white flowered parent plant (aa) produces (a) gametes.

After cross-pollination of parental plants, (A) and (a) unite together to form (Aa) zygote that develops into plants with pink flowers.

Here both the allelomorphic genes have a partial or incomplete dominant relationship and hence, F₁ hybrids show a mixture of characters of both parents. This is a case of incomplete dominance.

If allowed to self-pollinate, the F₁ pink blooming plants produce F₂ progeny bearing red, pink, and white flowers in a ratio of 1: 2 :1. This is the familiar genotypic ratio of an ordinary single gene F₁ self-cross.

What is new is that because the heterozygotes look unlike either homozygote, the phenotypic ratios are an exact reflection of the genotypic ratios.

F₂ Phenotypic ratio = 1 (Red): 2 (Pink): 1 (White) F₂ Genotypic ratio = 1 (AA): 2 (Aa): 1 (aa) In this example of Mirabilis jalapa, the red gene is incompletely dominant over the white gene and so both of them give rise to an intermediate pink colored flower in heterozygous or hybrid condition.

Explanation:

The biochemical explanation for this type of incomplete dominance is that each allele of the gene under analysis specifies an alternative form of a protein molecule with an enzymatic role in pigment production.

If the ‘white’ allele does not give rise to the functional enzyme, no pigment appears.

Thus, in Snapdragons and four o’clock, two ‘red’ alleles per cell produce a double dose of a red-producing enzyme, which generates enough pigment to make the flowers look fully red.

Wbbse Class 10 Life Science Heredity Notes

In the heterozygote, one copy of the ‘red’ allele per cell results in only enough pigment to make the flowers look pink. In the homozygote for the ‘white’ allele, where there is no functional enzyme and thus no red pigment, the flowers appear white.

WBBSE Chapter 3 Topic A Heredity Sex Determination In Human

The term sex refers to sexual phenotype. Most organisms have only two sexual phenotypes—male and female.

We, normally, define the sex of an individual organism about its phenotype. The mechanism by which sex is established is termed sex determination.

Sometimes an individual organism has chromosomes that are normally associated with one sex. For example, the cells of female humans normally have two X chromosomes, and the cells of males have one X chromosome and one Y chromosome.

Sex Determination In Humans:

In humans, Drosophila, and many other species, the cells of males and females have the same number of chromosomes, but the cells of males have a single X chromosome and a smaller sex chromosome, the Y chromosome.

The Y chromosome is not Y-shaped as is commonly assumed but is acrocentric. In this type of sex determination system, the male (44A + XY) is heterogametic because half of the male gametes have an X chromosome and the other half have a Y chromosome.

The female (44A + XX) is homogametic because all the eggs contain a single X chromosome. Fertilization of an egg (always X-bearing) with an X-bearing sperm produces female offspring (XX), but a Y-bearing sperm produces male offspring (XY).

The Total Chromosomes In Humans, In Each Body Cell, Can Be Represented As-

Female chromosomes – 44+XX where 44 are the autosomes and XX chromosomes are the sex chromosomes.

Male chromosomes – 44+XY where 44 are the autosomes and XY chromosomes are the sex chromosomes.

Although the X and Y chromosomes are not homologous, they pair and segregate into different cells in meiosis.

This is because of the fact that these chromosomes are homologous in small regions, called the pseudoautosomal region, in which they carry the same genes. In both types, the human X and Y chromosomes contain pseudoautosomal regions.

In humans and other placental mammals, maleness is due to a dominant effect of the Y chromosome. This is evidenced by the study of individuals with an abnormal number of sex chromosomes or aneuploidy.

XO persons (Turner syndrome) develop as females and XXY persons (Klinefelter syndrome) develop as males.

The dominant effect of the Y chromosome is exhibited early in development when it directs the primordial gonads to develop into testes.

Mendel’S Laws Of Heredity Class 10

Once the testes are formed, they secrete the hormone testosterone, which stimulates the development of male secondary sexual characteristics.

It is now known that the testis-determining factor (TDF) is the product of a gene called SRY (Sex-determining region Y), which is located outside the pseudoautosomal region in the short arm of the Y chromosome.

When fertilization occurs, the zygote (the initial cell from which a fetus grows) always inherits one of the mother’s X chromosomes, and either an X or a Y from the father, depending on which chromosome the fertilizing sperm cell happened to inherit.

One could say, then, that the father or, at least, his sperm determines the sex of the child.

The generally accepted theory is that males determine the sex because males can donate either an X chromosome or Y chromosome, while females can only donate an X chromosome to their offspring, making their contribution constant and the male’s contribution.

The variable, which under normal circumstances, determines the offspring’s genetic sex (at least, in humans Moreover, genetically there is a 50% chance of having a boy and a 50% chance of having a girl, as is found out from.

But there are actually slightly more boys born every year than girls.

It’s unclear why this is the case, but some research points out that more female fetuses die during pregnancy than male.

The Y chromosome contains all the directions that make the human zygote develop into a male. It is a relatively small chromosome with about 30 genes.

In comparison, the X chromosome has between 800 and 900 genes.

With its limited number, the Y chromosome focuses primarily on male traits. It contains the all-important SRY gene, which instructs the embryo to develop male traits such as testicles.

Another gene unique to the Y chromosome is USP9Y, which contributes to sperm production.

WBBSE Chapter 3 Topic A Heredity Fill In The Blanks