CBSE Class 10 Science Heredity And Evolution VBQs Set 04

This page contains 18 value based and competency based questions with answers for CBSE Class 10 Science Chapter 8, Heredity and Evolution, built around real-life situations such as ultrasound and female foeticide, blood group myths, earlobes in a family, a falling sex ratio and crops grown in different fields. They test how well you can apply Mendel's laws, Punnett squares and sex determination to everyday problems while thinking about social values, with each answer explained step by step.

Value Based Questions

Question 1. Mrs. Joshi, an eight months pregnant lady was suggested by her doctor to get an ultrasound done. She went to a radiologist with her husband and got the ultrasound done. When the ultrasound was done, her husband asked doctor about the sex of the baby in the womb.
(i) Is it ethical to determine the sex of the foetus? Why?
(ii) What is the chance of giving birth to a girl child in human beings?
(iii) What has government done to stop female foeticides?
Answer:
(i) No, it is not ethical. Finding out the sex of the foetus can lead to female foeticide, the killing of a girl child before birth. This is cruel and it upsets the balance of boys and girls in society, so the doctor should refuse to tell the sex.
(ii) The chance of a girl child is 50%, because an X sperm and a Y sperm fertilise the egg with equal chance.
(iii) The government has banned prenatal (before birth) sex determination tests and made sex-selective abortion a punishable offence, and it runs awareness campaigns to value the girl child.
In simple words: Telling the sex of the baby is banned because it can lead to female foeticide; every baby has a 50% chance of being a girl.

Teacher's Note:
1. Give the ethical reason: it can lead to female foeticide.
2. Mention the legal ban in part (iii).

Question 2. Ram met with an accident. John his school mate takes him to the hospital where Ram (AB blood group) needs blood transfusion. John also has AB blood group and is willing to donate his blood but Ram's mother objects by saying "John belongs to different community so has different type of blood". Give your opinion about Ram's mother views.
Answer:
Ram's mother's view is wrong.
A person's blood group is a genetic trait decided by the alleles inherited from the parents (A, B and O). It does not depend on community, religion or caste. People of every community can have the blood group AB.
Since John and Ram have the same blood group (AB), John can safely donate blood. Helping a person in need is the right thing to do, and one should think scientifically instead of believing such myths.
In simple words: Blood group is inherited through genes, not through community, so John can donate blood to Ram.

Teacher's Note:
1. Say that blood group is a genetically inherited trait.
2. Support the values of humanity and a scientific attitude.

Question 3. The given box diagram represents the ratio of females to males or the sex ratio in our country for 10 decades (1901 to 2001). Answer the following questions in the light of your knowledge of sex determination and the data presented in the box diagram.
(i) What does the bar diagram show?
(ii) As per scientific knowledge regarding sex determination, what should be the sex ratio or the male to female ratio at a given point of time?
(iii) Assign one reason to the trend showing deviation from the expected sex ratio.
(iv) Suggest a way by which such a trend can be stopped.
Answer:
(i) The bar diagram shows the number of females per thousand males in the population of India in each census year from 1901 to 2001. It fell from 972 in 1901 to 933 in 2001.
(ii) By sex determination, a child is equally likely to be a boy or a girl (50 : 50), so the male to female ratio should be 1 : 1, or about 1000 females per 1000 males.
(iii) Female foeticide (and neglect of the girl child) is the main reason for the deviation.
(iv) Banning and strictly enforcing the law against sex determination tests of the unborn baby, and spreading awareness and education to value the girl child.
In simple words: Fewer than 1000 girls per 1000 boys means girls are being lost, and the trend can be stopped by law and awareness.

Teacher's Note:
1. Read the axis carefully: it is females per thousand males.
2. Link the answer to equal chance of X and Y sperms.

Teacher's Note:
1. Use the table: 22 pairs of autosomes and 1 pair of sex chromosomes.
2. Explain why the father decides the sex.

Question 5. A husband has 46 chromosomes and his wife also has 46 chromosomes. Then why do not their offsprings have 46 pairs of chromosomes which are obtained by fusion of male and female gametes?
Answer:
Gametes (sperms and eggs) are formed by a special cell division, meiosis (reduction division), which halves the chromosome number. So each gamete has only 23 chromosomes, not 46.
When a sperm (23) fuses with an egg (23) at fertilisation, the zygote gets 23 + 23 = 46 chromosomes (23 pairs), not 46 pairs. Thus the chromosome number stays constant from one generation to the next.
In simple words: Each gamete brings only 23 chromosomes, so the child gets 46, which is 23 pairs.

Teacher's Note:
1. Do not confuse 46 chromosomes with 46 pairs.
2. Name meiosis as the reason for halving.

Question 6. A woman with blonde curly hair married a man with black soft hair. All of their children in first generation had black soft hair but in next generation children had different combinations in the ratio of 9 : 3 : 3 : 1. State the law that governs this expression.
Answer:
The law of independent assortment governs this expression.
All children in the first generation had black soft hair, so black and soft are dominant. In the next generation, four types appear in the ratio 9 : 3 : 3 : 1 (black soft : black curly : blonde soft : blonde curly). The new combinations show that the factors for hair colour and hair type separate during gamete formation and recombine independently of each other.
In simple words: A 9 : 3 : 3 : 1 ratio means the two traits are inherited independently.

Teacher's Note:
1. Identify the dominant traits from the first generation.
2. Name the law and give the reason.

Teacher's Note:
1. Divide the numbers: 60 : 20 = 3 : 1.
2. 3 : 1 comes only from Bb × Bb.

Question 8. The rules for inheritance of such traits in human beings are related to the fact that both the father and the mother contribute practically equal amounts of genetic material to the child. This means that each trait can be influenced by both paternal and maternal DNA.
(i) What is inheritance?
(ii) Who gave the three laws of inheritance?
(iii) What is the meaning of paternal and maternal DNA?
(iv) What is the phenotypic ratio of the dihybrid cross?
Answer:
(i) Inheritance is the passing of traits (characters) from the parents to their offspring, generation after generation.
(ii) Gregor Johann Mendel, who worked on garden pea plants, gave the laws of inheritance (dominance, segregation and independent assortment).
(iii) Paternal DNA is the DNA received from the father and maternal DNA is the DNA received from the mother.
(iv) The phenotypic ratio of a dihybrid cross in F2 is 9 : 3 : 3 : 1.
In simple words: A child gets equal DNA from both parents, and Mendel's dihybrid cross gives 9 : 3 : 3 : 1.

Teacher's Note:
1. Name Mendel and his pea plants.
2. Answer each part in one or two lines.

Real Life and Competency Based Questions

Question 9. Observe the ears of all the students in your class. The lowest part of the ear, called the earlobe, is closely attached to the side of the head in some of us, and not in others. Free and attached earlobes are two variants found in human populations. The percentage of people having free earlobes and attached earlobes is shown below:
(A) Based on the findings shown above what can be concluded?
(B) What will be the earlobes of a child whose parents have free earlobes?
(C) Mrs. and Mr. Sharma have six children. Three of them have attached earlobes like their father, and the other three have free earlobes like their mother. Taking F for free earlobes and f for attached earlobes, what genotype of parents and children?
Answer:
(A) Free earlobes are found in 84% of people and attached earlobes in only 16%, so free earlobes are the common form. This suggests that the trait for free earlobes is dominant and the trait for attached earlobes is recessive.
(B) A child of two parents with free earlobes may have either free or attached earlobes. If both parents are heterozygous (Ff), the child can be FF, Ff (free) or ff (attached).
(C) Children with attached earlobes are ff, so each of them got an f from both parents. The father has attached earlobes, so he is ff. The mother has free earlobes but the children include ff, so the mother must carry f, so she is Ff. The three children with attached earlobes are ff and the three with free earlobes are Ff.
In simple words: Free earlobes are dominant (F) and attached earlobes are recessive (ff); Ff × ff gives children of both kinds.

Teacher's Note:
1. Read the percentages from the pie chart.
2. In part (C) start from the recessive parent ff.

Teacher's Note:
1. Use the genotype table to read the phenotypes.
2. The F2 phenotype ratio is 3 white : 1 brown.

Question 11. Sex determination in humans happens through sex chromosomes. Along with other parameters, such processes often help in forensic studies in crime investigation and/or identification of accidents and natural calamities.
(A) In order to determine whether an accident victim is male or female, which cells can be used and why?
(B) Two pea plants one with round yellow seeds (RRYY) and another with wrinkled green (rryy) seeds produce F1 progeny that have round yellow (RrYy) seeds.
(i) When F1 plants are self pollinated, which new combination of characters is expected in F2 progeny?
(ii) What is the cause for the appearance of new combinations in F2 progeny?
Answer:
(A) Any body cell can be used, for example cells from blood, skin or hair roots, because every body cell has the sex chromosomes as the 23rd pair. XX shows a female and XY shows a male.
(B)(i) The new combinations in F2 are round green (Rryy or RRyy) and wrinkled yellow (rrYy or rrYY) seeds, besides the parental round yellow and wrinkled green. The F2 ratio is 9 : 3 : 3 : 1.
(ii) Seed shape and seed colour are inherited independently, since their factors assort independently during gamete formation, so the traits recombine in new ways.
In simple words: Any body cell shows the XX or XY pair; new seed combinations appear because the two traits are inherited independently.

Teacher's Note:
1. Say that all body cells carry the sex chromosomes.
2. Name the two new combinations.

Question 12. Mustard was grown in two fields- A and B. While field A produced brown coloured seeds, field B produced yellow coloured seeds. It was observed that in field A, the offsprings showed only the parental trait for consecutive generations, whereas in field B, majority of the offsprings showed a variation in the progeny.
What are the probable reasons for these?
Answer:
Field A: the plants reproduce by self-pollination, so the same genes are passed on and the offspring show only the parental trait generation after generation.
Field B: the plants undergo cross-pollination, so the genes of two different plants are recombined (and mixed during gamete formation), which gives variation among the offspring.
In simple words: Self-pollination keeps the same trait; cross-pollination mixes genes and gives variation.

Teacher's Note:
1. Name the type of pollination for each field.
2. Link variation to mixing of genes.

Question 13. What is the probability of a girl or a boy being born in a family? Justify your answer.
Answer:
There is a 50% chance of a girl and a 50% chance of a boy in every birth.
The mother has two X chromosomes, so all her eggs carry X. The father has X and Y, so half of his sperms carry X and half carry Y. A child who gets an X chromosome from the father is a girl (XX) and one who gets a Y chromosome is a boy (XY). Since the two kinds of sperms are equally likely to fertilise the egg, the chances of a girl and a boy are equal.
In simple words: X and Y sperms are equally likely, so a boy or a girl is equally likely.

Teacher's Note:
1. State the probability as 50 : 50.
2. Explain with the X and Y sperms.

Question 14. Pea plants can have green or yellow seeds. One of the phenotypes shows dominance over the other. A farmer decides to pollinate a flower of a plant with green seeds using pollen from a flower of a plant with yellow seeds. The resulting pod has all green seeds.
(A) What should be the genotype of the parent plants to give green and yellow seeds in equal proportions?
(B) The farmer crosses two heterozygous green seeded plants and obtains 100 plants in the F1 generation. What would be the number of green and yellow seeds respectively in the F1 generation?
Answer:
The cross gave all green seeds, so green (G) is dominant and yellow (g) is recessive.
(A) The genotypes of the parent plants should be Gg (green) × gg (yellow). This gives Gg and gg in equal numbers: 50% green and 50% yellow.
(B) Gg × Gg gives GG, Gg, Gg, gg, that is 3 green : 1 yellow. Out of 100 plants: 75 green and 25 yellow.
In simple words: A heterozygous green plant crossed with a yellow plant gives equal green and yellow; two hybrids give 75 : 25.

Teacher's Note:
1. Find the dominant trait from the first cross.
2. Convert the 3 : 1 ratio to numbers out of 100.

Case Based Questions

Question 15. A person first crossed pure-breed pea plants having round-yellow seeds with pure-breed pea plants having wrinkled-green seeds and found that only A-B type of seeds were produced in the F1-generation. When F1-generation pea plants having A-B type of seeds were cross-breed by self-pollination, then in addition to the original round yellow and wrinkled-green seeds, two new varieties A-D and C-B types of seeds were also obtained.
(i) What are A-B type of seeds?
(ii) State whether A and B are dominant traits or recessive traits.
(iii) What are A-D type of seeds?
(iv) What are C-B type of seeds?
(v) Out of A-B and A-D types of seeds, which one will be produced in (a) minimum number and (b) maximum number in the F2-generation?
Answer:
Round-yellow seeds are RY and wrinkled-green seeds are ry. In the F1 generation all seeds are round yellow, so A = round, B = yellow, C = wrinkled and D = green.
(i) A-B seeds are round yellow seeds.
(ii) A (round) and B (yellow) are dominant traits, since only they appear in F1.
(iii) A-D seeds are round green seeds.
(iv) C-B seeds are wrinkled yellow seeds.
(v) The F2 ratio is round yellow 9 : round green 3 : wrinkled yellow 3 : wrinkled green 1. (a) A-D (round green, 3 out of 16) is produced in the minimum number of the two; (b) A-B (round yellow, 9 out of 16) in the maximum number.
In simple words: Round yellow is 9/16 of F2 and the new combination round green is only 3/16.

Teacher's Note:
1. First decode A, B, C and D from the F1 result.
2. Use the 9 : 3 : 3 : 1 ratio for part (v).

Question 16. Observe the following cross between tall plants having round seeds and dwarf plants having wrinkled seeds. The individuals obtained in the F1-generation were thereafter self-crossed.
TTRR × ttrr
(Tall, Round) (Dwarf, Wrinkled)
TtRr × TtRr (F1-generation self-crossed)
(Tall, Round) (Tall, Round)
(i) What would be the phenotypes of the individuals obtained in the F2-generation? Give their ratios.
(ii) Why do you think all the individuals of the F1-generation were tall with round seeds?
(iii) What will be the number of progeny obtained in F2-generation in a dihybrid cross with pure dominant traits?
(iv) What would be the phenotype of the individual obtained in F1-generation?
(v) According to you, what is responsible for the inheritance of traits?
Answer:
(i) F2 phenotypes: tall round : tall wrinkled : dwarf round : dwarf wrinkled = 9 : 3 : 3 : 1.
(ii) Tallness and round seeds are dominant traits. Every F1 plant is TtRr, so the dominant alleles T and R hide t and r.
(iii) Out of the 16 combinations of the F2 square, only 1 is pure dominant for both traits (TTRR); 9 of the 16 show both dominant traits.
(iv) All F1 individuals are tall with round seeds.
(v) Genes (DNA on the chromosomes), which are passed from the parents to the offspring through the gametes, are responsible for the inheritance of traits.
In simple words: F2 shows tall round, tall wrinkled, dwarf round and dwarf wrinkled in the ratio 9 : 3 : 3 : 1.

Teacher's Note:
1. Quote the 9 : 3 : 3 : 1 ratio for part (i).
2. Part (iii) refers to the TTRR box of the F2 square.

Teacher's Note:
1. Experiment B shows the use of a test cross.
2. Compare 1 : 2 : 1 with 1 : 1.

Question 18. Most human chromosomes have a maternal and a paternal copy, and we have 22 such pairs. But one pair, called the sex chromosomes, is odd in not always being a perfect pair. Women have a perfect pair of sex chromosomes, both called X. But men have a mismatched pair in which one is a normal-sized X while the other is a short one called Y. So women are XX, while men are XY.
(i) If a normal cell of human body contains 46 chromosomes then the numbers of chromosomes in a sex cell of a human being is most likely to be: (a) 60 (b) 23 (c) 22 (d) 40
(ii) Which of the following determines the sex of a child? (a) The length of the mother's pregnancy (b) The length of time between ovulation and copulation (c) The presence of an X chromosome in an ovum (d) The presence of a Y chromosome in a sperm
(iii) The process where characteristics are transmitted from parent to offsprings is called: (a) Variation (b) Heredity (c) Gene (d) Allele
(iv) Who have a perfect pair of sex chromosomes? (a) Girls only (b) Boys only (c) Both girls and boys (d) It depends on many other factors
Answer:
(i) (b) 23. A sex cell (gamete) has half the number of chromosomes of a body cell, 46 ÷ 2 = 23.
(ii) (d) The presence of a Y chromosome in a sperm. Every egg carries X, so the sperm decides: a Y sperm gives a boy and an X sperm gives a girl. The length of pregnancy or timing has no effect.
(iii) (b) Heredity. The transmission of characters from parents to offspring is heredity.
(iv) (a) Girls only. Girls have two similar X chromosomes (XX); in boys the X and Y chromosomes are unequal.
In simple words: Gametes have 23 chromosomes; the father's sperm (X or Y) decides the sex; only girls have a matching XX pair.

Teacher's Note:
1. Gametes are haploid, so halve 46.
2. Boys have the unequal pair X and Y.