CBSE Class 10 Science Heredity VBQs Set 03

Accumulation of Variation During Reproduction

Question. In an asexually reproducing species, if a trait X exists in 5% of a population and trait Y exists in 70% of the same population, which of the two traits is likely to have arisen earlier? Give reason.
Answer: Trait Y is likely to have arisen earlier. In asexual reproduction, variations are few and occur only due to small inaccuracies in DNA copying. Once a variation or trait appears, it is inherited by the next generation and its frequency increases over time. Since trait Y exists in a much larger percentage (70%) of the population compared to trait X (5%), it indicates that trait Y has been present for many more generations and thus arose earlier.

Question. If a pure tall pea plant is crossed with a pure dwarf pea plant then, what percentage of \( F_1 \) and \( F_2 \) generation respectively will be tall?
(a) 25%, 25%
(b) 50%, 50%
(c) 75%, 100%
(d) 100%, 75%
Answer: (d) 100%, 75%

Question. Assertion (A) : Height in pea plants is controlled by efficiency of enzymes and is thus genetically controlled.
Reason (R) : Cellular DNA is the information source for making proteins in the cell.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true and (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false but (R) is true.
Answer: (a) Both (A) and (R) are true and (R) is the correct explanation of (A).

Question. Assertion (A) : A geneticist crossed a pea plant having violet flowers with a pea plant with white flowers, he got all violet flowers in first generation.
Reason (R) : White colour gene is not passed on to next generation.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (c) (A) is true, but (R) is false.

Question. Two pea plants - one with round yellow seeds (RRYY) and another with wrinkled green (rryy) seeds produce \( F_1 \) progeny that have round, yellow (RrYy) seeds. When \( F_1 \) plants are self-pollinated, which new combination of characters is expected in \( F_2 \) progeny? How many seeds with these new combinations of characters will be produced when a total 160 seeds are produced in \( F_2 \) generation? Explain with reason.
Answer: The new combinations of characters expected in the \( F_2 \) progeny are Round-Green and Wrinkled-Yellow. These are recombinant phenotypes that differ from the parental combinations (Round-Yellow and Wrinkled-Green).
In a dihybrid cross, the phenotypic ratio for \( F_2 \) is 9:3:3:1. The proportions for the new combinations are:
Round, green = \( \frac{3}{16} \)
Wrinkled, yellow = \( \frac{3}{16} \)
Total proportion of new combinations = \( \frac{3}{16} + \frac{3}{16} = \frac{6}{16} \)
Total number of seeds produced = 160
Number of seeds with new combinations = \( \frac{6}{16} \times 160 \)

\( \implies \) 60 seeds.
Reason: This result is obtained because of the Law of Independent Assortment, which states that different pairs of traits (seed shape and seed color) segregate and assort independently of each other during gamete formation.

Question. After self-pollination in pea plants with round yellow seeds, following types of seeds were obtained by Mendel:
Seeds | Number
Round, yellow | 630
Round, green | 216
Wrinkled, yellow | 202
Wrinkled, green | 64
Analyse the result and describe the mechanism of inheritance which explains these results.
Answer: Analysis of the result:
The ratio of the different types of seeds is approximately 630:216:202:64, which simplifies to approximately 9:3:3:1.
Mechanism of inheritance:
This result is explained by Mendel’s Law of Independent Assortment. It describes how different genes independently separate from one another when reproductive cells develop. In this case, the gene for seed shape (round/wrinkled) assorts independently of the gene for seed color (yellow/green). This leads to the formation of gametes with all possible combinations of alleles, resulting in parental types as well as new recombinant types in the \( F_2 \) generation.

Question. In humans, there is a 50% probability of the birth of a boy and 50% probability that a girl will be born. Justify the statement on the basis of the mechanism of sex-determination in human beings.
Answer: Human beings have 23 pairs of chromosomes, out of which one pair consists of sex chromosomes. Females have two identical sex chromosomes called X chromosomes (XX), while males have two different sex chromosomes, one X and one Y chromosome (XY).
During gamete formation, all eggs produced by a female will carry only one X chromosome. However, a male produces two types of sperm: 50% carry the X chromosome and 50% carry the Y chromosome.
- If a sperm carrying the X chromosome fertilizes an egg (X), the resulting zygote will be XX, developing into a girl.
- If a sperm carrying the Y chromosome fertilizes an egg (X), the resulting zygote will be XY, developing into a boy.
Since both types of sperm are produced in equal numbers, there is an equal (50%) probability for the child to be either male or female.

Pooja has green eyes while her parents and brother have black eyes. Pooja’s husband Ravi has black eyes while his mother has green eyes and father has black eyes.

Question. On the basis of the above given information, is the green eye colour a dominant or recessive trait? Justify your answer.
Answer: Green eye colour is a recessive trait. This is justified by the fact that Pooja has green eyes even though both of her parents have black eyes. This can only happen if both parents are heterozygous (carriers) of the recessive green-eye allele. If green eyes were a dominant trait, at least one of her parents would have had green eyes.

Question. What is the possible genetic makeup of Pooja’s brother’s eye colour?
Answer: Let 'B' represent the dominant allele for black eyes and 'b' represent the recessive allele for green eyes. Pooja's parents must be heterozygous (Bb) to have a green-eyed child (bb). Their children can have genotypes BB (Homozygous black) or Bb (Heterozygous black). Therefore, the possible genetic makeup of Pooja's brother, who has black eyes, is either BB or Bb.

Question. What is the probability that the offspring of Pooja and Ravi will have green eyes? Also, show the inheritance of eye colour in the offspring with the help of a suitable cross.
Answer: Pooja's genotype is bb (green eyes). Ravi's mother has green eyes (bb), so Ravi must have inherited a 'b' allele from her. Since Ravi has black eyes, his genotype must be Bb (heterozygous).
The cross between Pooja (bb) and Ravi (Bb):
Pooja (bb) x Ravi (Bb)
Gametes: (b) and (B, b)
Offspring genotypes: Bb (Black eyes), bb (Green eyes)
Ratio: 1:1
The probability that the offspring will have green eyes is 50% or \( \frac{1}{2} \).

Question. 50% of the offspring of Pooja’s brother are green eyed. With help of cross show how this is possible.
Answer: For 50% of the offspring to be green-eyed (bb), the brother must cross with a partner who is green-eyed (bb) or heterozygous (Bb), and the brother himself must be heterozygous (Bb).
Case: Brother (Bb) x Partner (bb)
Gametes: (B, b) and (b)
Offspring:
- Bb (Black eyes) - 50%
- bb (Green eyes) - 50%
This shows how it is possible for 50% of the offspring to have green eyes if the brother is a carrier (Bb) and his partner is green-eyed (bb).

Question. Sahil performed an experiment to study the inheritance pattern of genes. He crossed tall pea plants (TT) with short pea plants (tt) and obtained all tall plants in \( F_1 \) generation. What will be set of genes present in the \( F_1 \) generation?
Answer: The set of genes (genotype) present in the \( F_1 \) generation will be Tt (Heterozygous tall). This is because the \( F_1 \) plants receive one dominant allele (T) from the tall parent and one recessive allele (t) from the short parent.

Case Based Questions

Read the passage given below and answer the following questions :
In humans, the allele for brown eyes (B) is dominant over that for blue eyes (b). A brown eyed woman marries a blue eyed man, and they have six children. Four of the children are brown eyed and two of them are blue eyed.

Question. What is the genotype of blue eyed offspring?
(a) BB
(b) Bb
(c) bb
(d) Cannot be determined
Answer: (c) bb

Question. What is the woman’s genotype?
(a) BB
(b) Bb
(c) bb
(d) Cannot be determined
Answer: (b) Bb

Question. The ovum, produced by the mother carries the gene regarding eye colour is
(a) BB
(b) Bb
(c) B or b
(d) B only.
Answer: (c) B or b

Question. The ratio of brown eyed children to blue eyed children in this family is 2 : 1, which deviates from typical phenotypic ratios for monohybrid inheritance. What might be the reason?
(a) Gametes carrying the brown eyed allele are more viable than those with the blue eyed allele.
(b) A different pattern of inheritance other than monohybrid inheritance is involved.
(c) Not all of their babies survived childbirth, thus causing a distortion in the actual ratio.
(d) The actual ratio differs from the expected ratio because the sample size is too small.
Answer: (d) The actual ratio differs from the expected ratio because the sample size is too small.

Question. What is the gene carried by the man’s sperm regarding the eye colour?
(a) BB
(b) Bb
(c) b only
(d) b or B
Answer: (c) b only

Read the passage given below and answer the following questions :
Refer to the given table regarding results of \( F_2 \) generation of Mendelian cross.
Plants with round and yellow coloured seeds (P) : 315
Plants with round and green coloured seeds (Q) : 108
Plants with wrinkled and yellow coloured seeds (R) : 101
Plants with wrinkled and green coloured seeds (S) : 32

Question. Which of the following would be the phenotype of \( F_1 \) generation regarding given data of \( F_2 \) generation?
Answer: The phenotype of \( F_1 \) progeny is round and yellow seeds. This is because in a dihybrid cross, the \( F_2 \) generation shows a 9:3:3:1 ratio where the most frequent phenotype (Round and Yellow) represents the dominant traits expressed in \( F_1 \).

Question. Which of the following would be the genotype of parental generation regarding given result of \( F_2 \) generation?
Answer: The genotype of the parental generation is \( RRYY \times rryy \).

Question. If plant with wrinkled and green coloured seeds (S) is crossed with plant having wrinkled and yellow coloured seeds (R), what will be the probable phenotype of offsprings?
Answer: Plant with wrinkled and green coloured seeds (S) has genotype \( rryy \). Plant with wrinkled and yellow coloured seeds (R) can have genotype \( rrYY \) or \( rrYy \). If crossed with \( rrYY \), all offspring will be wrinkled and yellow (\( rrYy \)). If crossed with \( rrYy \), 50% offspring will be wrinkled and yellow (\( rrYy \)) and 50% will be wrinkled and green (\( rryy \)).

Question. What will be the phenotype of the offsprings, if RRYy is crossed with rrYy?
Answer: When \( RRYy \) is crossed with \( rrYy \), the offspring will be round yellow and round green. The cross results in genotypes \( RrYY \), \( RrYy \), and \( Rryy \) in a 1:2:1 ratio, which translates to a phenotypic ratio of 3 Round Yellow : 1 Round Green.

A & R Questions

In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following.
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.

Question. Assertion : The principle of segregation given by Mendel is the principle of purity of gametes.
Reason : Gametes are pure for a character and do not mix up.
Answer: (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.

Question. Assertion : In a monohybrid cross, offsprings of \( F_1 \) generation express dominant character.
Reason : Dominance occurs only in heterozygous state.
Answer: (c) Assertion is correct statement but reason is wrong statement.

Multiple Choice Questions

Question. The information source for making proteins in the cell is the
(a) chromosome
(b) DNA
(c) enzyme
(d) nucleus.
Answer: (b) DNA

Question. In which of the following animals, individuals can change sex?
(a) Snail
(b) Human
(c) Fruit fly
(d) All of the options
Answer: (a) Snail

Question. Which of the following ratios can prove law of independent assortment?
(a) 1 : 3 : 1
(b) 2 : 1 : 1
(c) 9 : 3 : 3 : 1
(d) 2 : 1
Answer: (c) 9 : 3 : 3 : 1

Question. What will be the percentage of purple stemmed plants in the \( F_2 \) generation, when the \( F_1 \) generation resulted due to cross breeding of green stemmed (GG) tomato plants with purple stemmed (gg) tomato plants, are self pollinated?
(a) 10%
(b) 25%
(c) 75%
(d) 50%
Answer: (b) 25%

VSA Type Questions

Question. How do variations favour survival of a species?
Answer: Variations increase the chances of survival of a species in a changing environment as some individuals may possess traits that allow them to adapt to new conditions.

Question. How many pairs of allelic characters did Mendel study in pea plant?
Answer: Mendel studied seven pairs of contrasting allelic characters in pea plants.

Question. Who is known as “Father of Genetics”?
Answer: Gregor Johann Mendel is known as the "Father of Genetics".

Question. State any two properties due to which Mendel selected pea plant.
Answer: (i) Pea plants have clearly visible contrasting characters. (ii) They have a short life cycle and are easy to grow and cross-breed.

SA I Type Questions

Question. Crossing of a pea plant with purple flower and pea plant with white flowers, produces 50 plants with only purple flowers. On selfing, the plants produced 470 plants with purple flowers and 160 with white flowers. Explain the genetic mechanism accounting for the above results.
Answer: The initial cross suggests that purple flower color is dominant over white. The \( F_1 \) generation is all purple (heterozygous). Upon self-pollination, the \( F_2 \) generation shows a phenotypic ratio of approximately 3 purple : 1 white (470:160 ≈ 2.9:1), which is consistent with Mendel's Law of Segregation for a monohybrid cross.

Question. “The chromosome number of the sexually reproducing parents and their offspring is the same.” Justify this statement.
Answer: During sexual reproduction, gametes are formed through meiosis, which reduces the chromosome number to half (haploid). When fertilization occurs, the fusion of two haploid gametes (sperm and egg) restores the original diploid chromosome number in the zygote.

Question. “Different species use very different strategies for determining the sex of their new born.” Justify this statement.
Answer: Sex determination can be genetic (e.g., XY system in humans) or environmental. For example, in some reptiles like turtles, the temperature at which eggs are incubated determines the sex of the offspring.

Question. What were the results of Mendel’s monohybrid cross?
Answer: In Mendel's monohybrid cross, the \( F_1 \) generation always expressed the dominant trait, and the \( F_2 \) generation showed a phenotypic ratio of 3 dominant : 1 recessive.

Question. State law of dominance.
Answer: The Law of Dominance states that in a heterozygote, one allele (dominant) will mask the presence of another allele (recessive) for the same characteristic.

Question. Why is the progeny always tall when a tall pea plant is crossed with a short pea plant?
Answer: Progeny is always tall because the 'tall' trait is dominant over the 'short' trait. In the \( F_1 \) generation, the presence of at least one dominant allele (T) results in the tall phenotype.

SA II Type Questions

Question. If we cross pure-breed tall (dominant) pea plant with pure-breed dwarf (recessive) pea plant we will get pea plants of \( F_1 \) generation. If we now self-cross the pea plant of \( F_1 \) generation, then we obtain pea plants of \( F_2 \) generation.
(a) What do the plants of \( F_1 \) generation look like?
(b) State the ratio of tall plants to dwarf plants in \( F_2 \) generation.
(c) State the type of plants not found in \( F_1 \) generation but appeared in \( F_2 \) generation, mentioning the reason for the same.
Answer: (a) All plants in the \( F_1 \) generation will be tall.
(b) The ratio of tall plants to dwarf plants in \( F_2 \) is 3 : 1.
(c) Dwarf plants are not found in \( F_1 \) but appear in \( F_2 \). This is because the recessive allele for dwarfness is masked in the heterozygous \( F_1 \) generation but segregates and pairs with another recessive allele in the \( F_2 \) generation.

Question. With the help of a flow chart explain in brief how the sex of a newborn is genetically determined in human beings. Which of the two parents, the mother or the father, is responsible for determination of sex of a child?
Answer: Sex is determined by the sex chromosomes. Females are XX and males are XY. Mother contributes an X chromosome to all gametes. Father contributes either X or Y.
Flow chart:
Parents: Mother (XX) \( \times \) Father (XY)
Gametes: (X), (X) \( \times \) (X), (Y)
Offspring: XX (Girl), XY (Boy)
The father is responsible for sex determination because he produces two types of gametes (X or Y).

Question. Write the basic features of mechanism of inheritance.
Answer: (i) Characters are controlled by genes. (ii) Genes come in pairs (alleles). (iii) One allele can be dominant over the other. (iv) Alleles segregate during gamete formation.

Question. Define the following terms : (i) Inheritance (ii) Heredity (iii) Trait (iv) Variations
Answer: (i) Inheritance: The process by which characters are passed from parent to progeny.
(ii) Heredity: The transmission of genetic characters from parents to offspring.
(iii) Trait: A specific characteristic of an organism.
(iv) Variations: The differences among the individuals of a species.

Question. An individual inherits different traits from his parents. On what basis classification of traits as dominant and recessive is done?
Answer: A trait is classified as dominant if it is expressed in the phenotype even when only one copy of the allele is present (heterozygous state). A trait is recessive if it is expressed only when two copies of the allele are present (homozygous state).

LA Type Questions

Question. Define Mendel’s experimental technique.
Answer: Mendel’s technique involved: (i) Selection of pure-breeding varieties with contrasting traits. (ii) Controlled cross-pollination of these varieties to produce \( F_1 \). (iii) Self-pollination of \( F_1 \) plants to produce \( F_2 \) and subsequent generations. (iv) Statistical analysis of the results to derive laws of inheritance.

Question. What are monohybrid and dihybrid crosses? Explain with the help of Punnett square.
Answer: A monohybrid cross involves a single pair of contrasting traits (e.g., plant height). A dihybrid cross involves two pairs of contrasting traits (e.g., seed shape and color). Punnett squares are used to predict the genotypes and phenotypes of the offspring by mapping all possible gamete combinations from both parents.

Question. (a) Work out a cross upto \( F_2 \) generation between two pure breed pea plants, one bearing violet flowers and the other white flowers.
(b) (i) Name this type of cross. (ii) State the different laws of Mendel that can be derived from such a cross.
Answer: (a) Parent: \( VV \text{ (Violet)} \times vv \text{ (White)} \)
\( \implies F_1: Vv \text{ (All Violet)} \)
\( \implies F_2: 1 VV, 2 Vv, 1 vv \text{ (3 Violet : 1 White)} \)
(b) (i) Monohybrid cross. (ii) Law of Dominance and Law of Segregation.

Question. A blue colour flower plant denoted by BB is cross bred with that of white colour flower plant denoted by bb.
(i) State the colour of flower you would expect in their \( F_1 \) generation plants.
(ii) What must be the percentage of white flower plants in \( F_2 \) generation if flowers of \( F_1 \) plants are self-pollinated?
(iii) State the expected ratio of the genotypes BB and Bb in the \( F_2 \) progeny.
Answer: (i) All \( F_1 \) flowers will be blue (Bb).
(ii) In \( F_2 \), 25% of the plants will be white (bb).
(iii) The ratio of BB : Bb in \( F_2 \) is 1 : 2.

Question. Explain the following statements with an example.
(i) Traits may be dominant or recessive.
(ii) Inheritance of two traits is independent of each other.
Answer: (i) For example, in pea plants, tallness (T) is dominant and dwarfness (t) is recessive. A Tt plant is tall.
(ii) For example, in a dihybrid cross of round yellow and wrinkled green seeds, the inheritance of seed shape does not affect the inheritance of seed color, resulting in new combinations like round green and wrinkled yellow. This is the Law of Independent Assortment.