Question 1

Match List I with List II : List I List II A. Incomplete dominance I. Human skin colour B. Co-dominance II. Inheritance of flower colour in Antirrhinum sp. C. Pleiotropy III. Phenylketonuria disease i

Accepted Answer

Incomplete dominance is observed in the inheritance of flower colour in Antirrhinum sp. (snapdragon), where the heterozygous phenotype is intermediate between the two homozygous phenotypes. Co-dominance is exhibited by ABO blood groups in humans, where both alleles I^A and I^B express themselves fully in a heterozygous individual. Pleiotropy occurs when a single gene influences multiple phenotypic traits. Phenylketonuria is an example of pleiotropy in humans, where a single gene mutation leads t

Question 2

The sixth mutant codon of beta globin gene causing polymerization of Haemoglobin and change in RBC shape is :

Accepted Answer

Sickle cell anaemia is caused by the substitution of Glutamic acid (Glu) by Valine (Val) at the sixth position of the beta globin chain of the haemoglobin molecule. This substitution of the amino acid occurs due to a single base substitution at the sixth codon of the beta globin gene from GAG to GUG. The normal codon is GAG, which codes for Glutamic acid, while the mutant codon is GUG, which codes for Valine. The presence of Valine leads to the polymerization of the mutant haemoglobin molecule u

Question 3

In a population of a grasshopper species, the chromosome number of some members is 23 and some other members possess 24 chromosomes. The 23 and 24 chromosome-bearing members in this species are :

Accepted Answer

Grasshoppers exhibit the XX-XO type of sex determination. In this system, females have a pair of X chromosomes (XX) along with autosomes, while males have only one X chromosome (XO) along with autosomes. Thus, females have one more chromosome than males. Given the chromosome numbers 23 and 24, the members with 23 chromosomes are males (22 autosomes + 1 X chromosome) and the members with 24 chromosomes are females (22 autosomes + 2 X chromosomes). Therefore, the 23 and 24 chromosome-bearing membe

Question 4

Which one of the following disorders is caused by the substitution of Glutamic acid (Glu) by Valine (Val) at the sixth position of the beta globin chain of the haemoglobin molecule ?

Accepted Answer

Sickle-cell anaemia is an autosomal recessive genetic disorder caused by a point mutation in the DNA sequence of the gene encoding the beta globin chain of haemoglobin. This mutation results in the substitution of Glutamic acid (Glu) by Valine (Val) at the sixth position of the beta globin chain. The codon GAG (for Glutamic acid) is replaced by GUG (for Valine). Thalassemia is a quantitative defect in globin chain synthesis, Phenylketonuria is an inborn error of metabolism affecting phenylalanin

Question 5

What is the probability of having children with 'O' blood group, where both mother and father are heterozygous for 'A' and 'B' blood group, respectively ?

Accepted Answer

The genotype of the mother, who is heterozygous for 'A' blood group, is I^ A i. The genotype of the father, who is heterozygous for 'B' blood group, is I^ B i. A cross between the mother and father (I^ A i I^ B i) produces the following possible genotypes in the offspring: I^ A I^ B (AB blood group) I^ A i (A blood group) I^ B i (B blood group) ii (O blood group) Out of the four possible combinations, only one results in the 'O' blood group (ii). The probability of having a child with 'O' blood

Question 6

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R . Assertion A : In an experiment, Mendel observed that the F1 progeny plants are all tall and none

Accepted Answer

Assertion A is correct because when Mendel crossed true-breeding tall pea plants with true-breeding dwarf pea plants, all the plants in the F1 generation were tall. Reason R is correct and explains Assertion A. Stem height is a contrasting trait in pea plants where the tall trait is dominant over the dwarf trait. Since the F1 progeny are heterozygous, the dominant allele expresses itself, resulting in all tall plants. Therefore, both A and R are correct and R is the correct explanation of A. Ans

Question 7

Given below are two statements : Statement I : Down's syndrome is caused by the absence of one of the X-chromosomes. Statement II : Turner's syndrome is caused by the presence of an additional copy of

Accepted Answer

Down's syndrome is an autosomal aneuploidy caused by the presence of an additional copy of chromosome number 21 (Trisomy 21). Therefore, Statement I is incorrect. Turner's syndrome is a sex chromosomal aneuploidy caused by the absence of one of the X-chromosomes, resulting in a karyotype of 45 with XO. Therefore, Statement II is also incorrect. Both Statement I and Statement II are incorrect. Answer: Both Statement I and Statement II are incorrect

Question 8

Which of the following statements are true with reference to the sex-determination in honeybees ? A. An offspring formed from the union of a sperm and an egg, develops as a female (queen or worker). B

Accepted Answer

In honeybees, sex determination is based on the haplodiploid system. Statement A is correct. Fertilized eggs (diploid) develop into females, which can be either queens or workers depending on their nutrition. Statement B is correct. Unfertilized eggs develop into males (drones) through parthenogenesis. Statement C is correct. Males are haploid (n=16) and females are diploid (2n=32), so males have half the number of chromosomes compared to females. Statement D is incorrect. Since males are alread

Question 9

For a person with blood group ‘O’, which of the following is not a possible combination of parents' blood group genotypes ?

Accepted Answer

Blood group 'O' corresponds to the homozygous recessive genotype ii. For a child to have blood group 'O', they must inherit one i allele from each parent. This means both parents must possess at least one i allele in their genotype. In the combination Father : I^A I^B and Mother : I^A i, the father does not have an i allele. Therefore, he can only pass either an I^A or an I^B allele to the offspring. As a result, it is impossible for this couple to produce a child with the ii genotype (blood gro

Question 10

With the help of given pedigree, find out the probability for the birth of a child having no disease and being a carrier (has the disease mutation in one allele of the gene) in F_3 generation.

Accepted Answer

As in the F_1 generation the carrier female and non-affected (normal, not carrier) had affected male child that means the genetic disorder is sex-linked recessive. The consanguineous mating between female (X^C X ) and male (X^C Y ) Out of 4 child only one is carrier i.e. 1 4 .

Question 11

Genes R and Y follow independent assortment. If RRYY produce round yellow seeds and rryy produce wrinkled green seeds, what will be the phenotypic ratio of the F2 generation?

Accepted Answer

A classical dihybrid cross performed by Mendel involves. A cross which was made between a pure round yellow seeded pea plant (RRYY) with wrinkled green seeded plant (rryy). Yellow colour is dominant over green and round seed shape over wrinkled seed shape. Phenotypic ratio in F2 generation

Question 12

What is the pattern of inheritance for polygenic trait?

Accepted Answer

Polygenic inheritance refers to the inheritance of a trait controlled by two or more genes. When human disorders are determined by mutation in the single gene then they are transmitted to the offspring as per Mendelian principle. Polygenic trait shows non-Mendelian inheritance pattern.

Principles of Inheritance and Variation — Previous Year Questions & Solutions