NEET Biology Principles of Inheritance and Variation MCQs Set 14

Biology Objective Questions and Answers: Principles of Inheritance and Variation

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Multiple Choice Questions

Question 1: Read the following statements and choose the correct option.
(a) In phenylketonuria the affected person does not secrete the enzyme to convert phenylalanine to tyrosine.
(b) Possibility of male becoming haemophilic is extremely rare.
(c) Sickle cell anaemia is caused by the substitution of glutamic acid by valine at fifth position of beta chain of haemoglobin.
(d) Myotonic dystrophy is an X-linked recessive disease
Show Answer & Explanation

Answer: (a) In phenylketonuria the affected person does not secrete the enzyme to convert phenylalanine to tyrosine.

Explanation:
1. In phenylketonuria the enzyme that converts phenylalanine to tyrosine is missing.
2. Haemophilia is rare in females, not males; the sickle cell substitution is at the sixth position of the beta chain; myotonic dystrophy is autosomal dominant.

Teacher's Note:
1. PKU: phenylalanine hydroxylase deficiency.
2. Sickle cell: Glu to Val at position 6.

Question 2: A mother has Down's syndrome (trisomy 21) and marries a normal man. What is the probability that a child of this couple has Down's syndrome?
(a) 100%
(b) 75%
(c) 50%
(d) 25%
Show Answer & Explanation

Answer: (c) 50%

Explanation:
1. A woman with trisomy 21 produces gametes with 23 or 24 chromosomes in equal numbers.
2. So half of the offspring from a normal father are affected.

Teacher's Note:
1. Down syndrome mothers can have children.
2. Risk is 50 percent.

Question 3: The ability of certain people to taste the chemical phenylthiocarbamide (PTC) is governed by a dominant allele T and the inability to taste PTC by its recessive allele t. If 16% of a population is homozygous taster and 48% is heterozygous taster, what is the frequency of t?
(a) 0.3
(b) 0.9
(c) 0.4
(d) 0.6
Show Answer & Explanation

Answer: (d) 0.6

Explanation:
1. 16 percent are TT and 48 percent are Tt, so tt = 100 - 64 = 36 percent.
2. Frequency of t = square root of 0.36 = 0.6.

Teacher's Note:
1. Use Hardy-Weinberg: q squared = 0.36.
2. So q = 0.6.

Question 4: A man with normal vision whose father was colourblind marries with women whose father was also colourblind. Suppose their first child is daughter then what are the chances of this child to be colourblind?
(a) 100%
(b) 25%
(c) 50%
(d) 0%
Show Answer & Explanation

Answer: (d) 0%

Explanation:
1. The man is X^C Y and the woman is a carrier X^C X^c (father colour blind).
2. A daughter always receives the normal X^C from her father, so she cannot be colour blind.

Teacher's Note:
1. Girls need an affected X from both parents.
2. The father here is normal.

Question 5: Two genes A and B present on same chromosome show 20% recombination, then what will be the percentage of gamete (Ab) for the plant having genotype AaBb in cis configuration?
(a) 40%
(b) 80%
(c) 20%
(d) 10%
Show Answer & Explanation

Answer: (d) 10%

Explanation:
1. With 20 percent recombination, the recombinant gametes (Ab and aB) total 20 percent.
2. The plant is in cis form (AB/ab), so Ab is 10 percent.

Teacher's Note:
1. Cis: AB and ab are parental.
2. Ab is half of the recombinants.

Question 6: Phenotypically females having rudimentary ovaries, underdeveloped breasts, short stature, webbing neck, often subnormal intelligence suggests:
(a) Down's syndrome
(b) Klinefelter's syndrome
(c) Turner's syndrome
(d) Haemophilic syndrome
Show Answer & Explanation

Answer: (c) Turner's syndrome

Explanation:
1. Turner's syndrome (45, XO) gives phenotypic females with rudimentary ovaries, short stature, webbed neck and poorly developed breasts.
2. Klinefelter's is XXY male.

Teacher's Note:
1. Turner: XO.
2. Klinefelter: XXY.

Question 7: A test cross of F1 Drosophila shows following results:- ++ / ab  82 +a / ab  18 +b / ab  18 ab / ab  82 Select the incorrect statement for this result:-
(a) Parents have Cis arrangement of linked genes.
(b) Both genes are 18 cM apart on chromosome.
(c) Recombinant offspring have original arrangement of linked genes.
(d) The results indicate that coupling occurs between genes, in parents.
Show Answer & Explanation

Answer: (c) Recombinant offspring have original arrangement of linked genes.

Explanation:
1. The parental types are ++ and ab (82 each) and recombinants are +a and +b (18 each).
2. Recombination frequency is 36/200 = 18 percent, so the genes are 18 cM apart. Statement C is wrong because recombinants have new combinations of genes, not the original ones.

Teacher's Note:
1. Parents were in coupling (cis).
2. Recombinants have new arrangements.

Question 8: Given below are some sex- determination methods, which of the following methods found in arthropods. (a) xx-xy method (b) xx-xo method (c) zw-zz method (d) Haplodiploidy
(a) b and c
(b) a, b and c
(c) b, c and d
(d) All
Show Answer & Explanation

Answer: (d) All

Explanation:
1. Arthropods show XX-XY (Drosophila), XX-XO (grasshopper), ZW-ZZ (butterflies and moths) and haplodiploidy (honeybee).
2. So all methods are found.

Teacher's Note:
1. Haplodiploidy: bees.
2. ZW: Lepidoptera.

Question 9: Gene A and B are linked and are 24 cM apart on chromosome. An offspring of parents AAbb and aaBB will produces what percent of various gametes?
(a) 38% AB, 12% Ab, 12% aB, 38% ab.
(b) 12% AB, 38 Ab, 38% aB, 12% ab.
(c) 24% AB, 26% Ab, 26% aB, 24% ab.
(d) 26% AB, 24% Ab, 24% aB, 26% ab.
Show Answer & Explanation

Answer: (b) 12% AB, 38 Ab, 38% aB, 12% ab.

Explanation:
1. The parents AAbb and aaBB give F1 with Ab on one chromosome and aB on the other.
2. Parental gametes Ab and aB are 38 percent each; the recombinants AB and ab are 12 percent each (24 percent recombination).

Teacher's Note:
1. Recombinants = map distance.
2. Parentals = 100 - map distance.

Question 10: Determine the probability of genotype, MMNn; in the offspring of selfing dihybrid parents (MmNn):
(a) 1/4
(b) 1/8
(c) 1/16
(d) 1/2
Show Answer & Explanation

Answer: (b) 1/8

Explanation:
1. MmNn x MmNn gives MM in 1/4 of the offspring and Nn in 1/2.
2. The probability of MMNn is 1/4 x 1/2 = 1/8.

Teacher's Note:
1. Multiply the single gene probabilities.
2. Dihybrid: 16 combinations.

Question 11: Match List I with List II and select the correct answer.
List I: P. Incomplete dominance, Q. Codominance, R. Polygenes, S. Complementary genes
List II: 1. Human skin colour, 2. Purple colour in maize due to anthocyanin, 3. Human beings with AB blood group, 4. Pink flower in the 4 O'clock plant
(a) P-4, Q-3, R-1, S-2
(b) P-3, Q-4, R-2, S-1
(c) P-4, Q-3, R-2, S-1
(d) P-3, Q-4, R-1, S-2
Show Answer & Explanation

Answer: (a) P-4, Q-3, R-1, S-2

Explanation:
1. Incomplete dominance: pink flowers in the 4 o'clock plant. Codominance: AB blood group. Polygenes: human skin colour. Complementary genes: purple colour in maize due to anthocyanin.

Teacher's Note:
1. Different gene interactions.
2. Match each with the example.

Question 12: Which of the following feature is not associated with maternal inheritance?
(a) It is mostly uniparental
(b) It occurs due to nuclear genes
(c) Organelles genome is responsible for maternal inheritance
(d) Variegation in four O-clock plants is an example of maternal inheritance.
Show Answer & Explanation

Answer: (b) It occurs due to nuclear genes

Explanation:
1. Maternal inheritance is due to organelle (mitochondrial, chloroplast) genes and is mostly uniparental.
2. It is not due to nuclear genes, so B is not associated.

Teacher's Note:
1. Variegation in Mirabilis is plastid inherited.
2. Organelle genes follow the mother.

Question 13: If two dihybrid plants (AaBb) are crossed, find the correct combination of probabilities of the following offspring: AaBb, AABb, AaBB, aabb.
(a) AaBb 8/16, AABb 2/16, AaBB 2/16, aabb 1/16
(b) AaBb 4/16, AABb 2/16, AaBB 2/16, aabb 2/16
(c) AaBb 1/4, AABb 1/8, AaBB 1/8, aabb 1/8
(d) AaBb 1/4, AABb 1/8, AaBB 1/8, aabb 1/16
Show Answer & Explanation

Answer: (d) AaBb 1/4, AABb 1/8, AaBB 1/8, aabb 1/16

Explanation:
1. Offspring of AaBb x AaBb: AaBb = 4/16 = 1/4, AABb = 2/16 = 1/8, AaBB = 2/16 = 1/8 and aabb = 1/16.
2. This matches option 4.

Teacher's Note:
1. Use the 16 box checkerboard.
2. Count the genotypes.

Question 14: In first experiment, a cross was made between two plants A and B, in which A plant was gone through emasculation. Phenotypic result of F2 generation was 3: 1. In second experiment B plant was gone through emasculation. Result of second experiment should be (with respect to Mendelian traits):
(a) Same phenotypic ratio 3: 1
(b) Changed phenotypic ratio 1: 1
(c) Changed phenotypic ratio 3: 1
(d) Changed genotypic ratio 3: 1
Show Answer & Explanation

Answer: (a) Same phenotypic ratio 3: 1

Explanation:
1. Mendel's reciprocal crosses give the same result, because the pollen and egg contribute equally to the offspring for nuclear traits.
2. So the F2 ratio remains 3 : 1.

Teacher's Note:
1. Reciprocal crosses agree.
2. This confirms nuclear inheritance.

Question 15: When both alleles of a pair are fully expressed in a heterozygous, they are called:
(a) Lethals
(b) Co-Dominant
(c) Recessive alleles
(d) In-complete dominant
Show Answer & Explanation

Answer: (b) Co-Dominant

Explanation:
1. When both alleles are fully expressed in a heterozygote the alleles are codominant.
2. Incomplete dominance gives a blend.

Teacher's Note:
1. Example: AB blood group.
2. Both IA and IB act.

Question 16: When we make a cross between AAbb and aaBB then find out the phenotypic ratio we obtained in next generation:
(a) 3: 1
(b) 9: 3: 3: 1
(c) 1: 1
(d) 1: 0
Show Answer & Explanation

Answer: (d) 1: 0

Explanation:
1. AAbb x aaBB gives only AaBb in the next generation, all with the same phenotype.
2. So the ratio is 1 : 0.

Teacher's Note:
1. F1 is uniform.
2. 9:3:3:1 is seen in F2.

Question 17: In a dihybrid cross, if we get 9: 3: 3: 1 ratio it denotes that:
(a) The alleles of two genes are interacting with each other.
(b) It is a multigenic inheritance
(c) It is a case of multiple allelism
(d) The alleles of two genes are assorting independently
Show Answer & Explanation

Answer: (d) The alleles of two genes are assorting independently

Explanation:
1. A 9 : 3 : 3 : 1 ratio in a dihybrid cross shows independent assortment of the two genes.
2. Gene interaction changes this ratio.

Teacher's Note:
1. Mendel's second law.
2. Genes on different chromosomes.

Question 18: If the minimum and maximum height of sugarcane plant is 50 cm and 200 cm respectively and this character is regulated by 3 pairs of genes. Then find out the height of plants produced in the progeny of cross aabbCC × AABBcc:
(a) 75 cm
(b) 150 cm
(c) 50 cm
(d) 125 cm
Show Answer & Explanation

Answer: (d) 125 cm

Explanation:
1. Each contributing allele adds (200 - 50)/6 = 25 cm.
2. The cross gives AaBbCc with three capital alleles, so height = 50 + 3 x 25 = 125 cm.

Teacher's Note:
1. Polygenic: additive.
2. Count the capital letters.

Question 19: One of the parent of a cross has a mutation in its mitochondria. In that cross, that parent is taken as a male. During segregation of F2 progenies that mutation is found in:
(a) One third of the progenies
(b) None of the progenies
(c) All the progenies
(d) Fifty percent of the progenies
Show Answer & Explanation

Answer: (b) None of the progenies

Explanation:
1. Mitochondrial genes are inherited only from the mother.
2. The male parent with the mutation passes it to none of the progeny.

Teacher's Note:
1. Cytoplasmic inheritance.
2. Sperm gives no mitochondria.

Question 20: In a dihybrid cross the characters show complete dominance and incomplete dominance respectively. The F2 phenotypic ratio on selfing will be
(a) 9 : 3 : 3 : 1
(b) 3 : 1
(c) 3 : 6 : 3 : 1 : 2 : 1
(d) None of the above
Show Answer & Explanation

Answer: (c) 3 : 6 : 3 : 1 : 2 : 1

Explanation:
1. Complete dominance for one gene gives 3 : 1 and incomplete dominance for the other gives 1 : 2 : 1.
2. Combined, F2 ratio is 3 : 6 : 3 : 1 : 2 : 1.

Teacher's Note:
1. Multiply the two ratios.
2. Total 16 parts.

Question 21: In an organism, tall phenotype is dominant over recessive dwarf phenotype, and the alleles are designated as T and t, respectively. Upon crossing two different individuals, total 250 offspring were obtained, out of which 124 displayed tall phenotype and rest were dwarf. Thus, the genotype of the parents were:
(a) TT × TT
(b) TT × tt
(c) Tt × Tt
(d) Tt × tt
Show Answer & Explanation

Answer: (d) Tt × tt

Explanation:
1. About half the offspring are tall (124 of 250) and half are dwarf.
2. This 1 : 1 ratio is typical of a test cross, so Tt x tt.

Teacher's Note:
1. 1:1 means test cross.
2. Tt x tt.

Question 22: Two pink flowered snapdragon plants (Rr) are self pollinated. The probability of the offspring to have white flowers are:
(a) 25%
(b) 50%
(c) 75%
(d) 2.5%
Show Answer & Explanation

Answer: (a) 25%

Explanation:
1. Rr x Rr gives RR : Rr : rr in 1 : 2 : 1, and rr is white.
2. So 25 percent are white.

Teacher's Note:
1. Pink: Rr in snapdragon.
2. Incomplete dominance.

Question 23: The ABO blood grouping in human beings is an example of: (i) Dominance (ii) Incomplete dominance (iii) Co-dominance (iv) Multiple alleles
(a) (i) and (ii) only
(b) (ii), (iii) and (iv)
(c) (i), (iii) and (iv)
(d) (iii) and (ii) only
Show Answer & Explanation

Answer: (c) (i), (iii) and (iv)

Explanation:
1. ABO blood groups show dominance (IA, IB over i), codominance (IA and IB) and multiple alleles (three alleles).
2. It does not show incomplete dominance.

Teacher's Note:
1. Three alleles, six genotypes.
2. Four phenotypes.

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