NEET Biology Principles of Inheritance and Variation MCQs Set 09

Practice MCQs for NEET Biology Principles of Inheritance and Variation

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Multiple Choice Questions

Question 1: When a pink flowered Antirrhinum plant is test crossed, then phenotypic ratio in resulting progenies is
(a) 1 Red : 1 White
(b) 3 Red : 1 White
(c) 2 Pink : 1 White
(d) 1 Pink : 1 White
Show Answer & Explanation

Answer: (d) 1 Pink : 1 White

Explanation:
1. Pink Antirrhinum is Rr. Test cross Rr x rr gives Rr (pink) and rr (white) in 1 : 1.

Teacher's Note:
1. In incomplete dominance the test cross ratio is 1 : 1 for both phenotype and genotype.

Question 2: Heterozygous tall and violet flowered pea plants were selfed and total 512 seeds are collected. What will be total number of seeds for both heterozygous traits?
(a) 128
(b) 256
(c) 384
(d) 64
Show Answer & Explanation

Answer: (a) 128

Explanation:
1. TtRr selfed gives 16 combinations, of which TtRr is 4/16. So 512 x 1/4 = 128 seeds.

Teacher's Note:
1. Double heterozygote frequency in F2 is 1/4.

Question 3: Mark the odd one (w.r.t. F2 generation of Mendelian dihybrid cross)
(a) Frequency of TtRR genotype = 12.5%
(b) Frequency of ttrr genotype = 6.25%
(c) Frequency of TTRR genotype = 6.25%
(d) Frequency of ttRr genotype = 25%
Show Answer & Explanation

Answer: (d) Frequency of ttRr genotype = 25%

Explanation:
1. In an F2 dihybrid, ttRr is 2/16 = 12.5%, not 25%. TtRR is 12.5%, ttrr and TTRR are 6.25% each.

Teacher's Note:
1. Heterozygous for one gene and homozygous for the other is 2/16.

Question 4: Find out the frequency of AabbCcDdee if parents are AabbCCddEe and AabbccDdee
(a) 0.78%
(b) 12.5%
(c) 25%
(d) 50%
Show Answer & Explanation

Answer: (b) 12.5%

Explanation:
1. Aa x Aa gives Aa 1/2; bb x bb gives bb 1; CC x cc gives Cc 1; dd x Dd gives Dd 1/2; Ee x ee gives ee 1/2. Product = 1/2 x 1/2 x 1/2 = 1/8 = 12.5%.

Teacher's Note:
1. Multiply the probabilities of independent gene pairs.

Question 5: In incomplete dominance
(a) Dominant trait is completely expressed in F1 generation
(b) Phenotypic and genotypic ratio are different
(c) Two dominant alleles are needed to express the complete dominant trait
(d) F1 individuals have the equal traits of both parents
Show Answer & Explanation

Answer: (c) Two dominant alleles are needed to express the complete dominant trait

Explanation:
1. In incomplete dominance the heterozygote is intermediate, and the full dominant trait appears only with two copies of the dominant allele. F1 is not dominant, and phenotype and genotype ratios are the same (1:2:1).

Teacher's Note:
1. Compare with codominance, where both alleles express.

Question 6: Progeny with blood group ‘O’ can not be obtained in cross
(a) A×A
(b) A×B
(c) O×AB
(d) B×B
Show Answer & Explanation

Answer: (c) O×AB

Explanation:
1. O x AB gives IAi and IBi only (A and B groups). Groups O needs ii, which cannot arise.

Teacher's Note:
1. AB parents can never have O children.

Question 7: Which of the following parental combination has produced mutant offspring?
(a) Tt × tt = Tt
(b) tt × tt = Tt
(c) Tt × Tt = tt
(d) TT × tt =Tt
Show Answer & Explanation

Answer: (b) tt × tt = Tt

Explanation:
1. Tt can arise from tt x tt only if one t mutates to T, so that cross gives a mutant offspring.

Teacher's Note:
1. Crossing identical homozygotes should give identical offspring.

Question 8: Which of the following combination seems to have some linkage in character selected by Mendel?
(a) Stem height and pod colour
(b) Flower colour and flower position
(c) Seed shape and seed colour
(d) Plant height and pod shape
Show Answer & Explanation

Answer: (d) Plant height and pod shape

Explanation:
1. Plant height and pod shape are on the same chromosome (chromosome 4) and show some linkage, although it was not noticed by Mendel.

Teacher's Note:
1. Mendel's genes were on 4 chromosomes.

Question 9: A diploid organism is heterozygous for five loci and homozygous for 2 loci, how many types of gametes can be produced?
(a) 128
(b) 32
(c) 4
(d) 14
Show Answer & Explanation

Answer: (b) 32

Explanation:
1. Only the five heterozygous loci produce different gametes, so 2^5 = 32 gamete types.

Teacher's Note:
1. Homozygous loci do not add gamete types.

Question 10: Lesch Nyhan disease is an X-linked recessive disorder that causes neurological damage in human beings. A survey of 500 males from a Caucasian population revealed that 20 were affected with this disorder. What is the frequency of the normal allele in this population?
(a) 0.96
(b) 0.8
(c) 0.04
(d) 0.2
Show Answer & Explanation

Answer: (a) 0.96

Explanation:
1. The frequency of a recessive X-linked allele in males equals the disease frequency: q = 20/500 = 0.04. So the normal allele frequency p = 1 - 0.04 = 0.96.

Teacher's Note:
1. Males are hemizygous, so allele frequency equals male phenotype frequency.

Question 11: How many types of zygotic combinations are possible between a cross Aa BB Cc Dd × AA bb Cc DD?
(a) 32
(b) 128
(c) 64
(d) 16
Show Answer & Explanation

Answer: (d) 16

Explanation:
1. Parent 1 (AaBBCcDd) forms 2 x 1 x 2 x 2 = 8 gamete types; parent 2 (AAbbCcDD) forms 2. Zygotic combinations = 8 x 2 = 16.

Teacher's Note:
1. Multiply gamete types of both parents.

Question 12: Morgan hybridised yellow-bodied, white-eyed females to brown-bodied, red-eyed males and intercrossed their F1 progeny. He observed that
(a) F2 ratio was deviated very significantly from the 9 : 3 : 3 : 1 ratio
(b) Both genes did not segregate independently of each other
(c) Recombinant types are not obtained in F2 generation
(d) Both genes segregate independently of each other Select the correct set of statements :
(a) (a) & (b) only
(b) (b) & (c) only
(c) (b) & (d) only
(d) (c) & (d) only
Show Answer & Explanation

Answer: (a) (a) & (b) only

Explanation:
1. Yellow-body white-eye and brown-body red-eye genes are linked, so F2 deviates from 9:3:3:1 and the genes do not assort independently; recombinants still appear (low frequency).

Teacher's Note:
1. Linkage reduces recombinants.

Question 13: ___(A)___ used the frequency of recombination between gene pairs on the ___(B)___ as a measure of the distance between genes and mapped their position on the chromosome.
(A) (B)
(a) Morgan Same chromosome
(b) Sturtevant Different chromosomes
(c) Morgan Different chromosomes
(d) Sturtevant Same chromosome
Show Answer & Explanation

Answer: (d) Sturtevant Same chromosome

Explanation:
1. Sturtevant used the recombination frequency between gene pairs on the same chromosome to measure the distance between them and mapped genes.

Teacher's Note:
1. Morgan discovered linkage; Sturtevant made maps.

Question 14: While solving the problem of sex determination in large number of insects, it was observed that
(a) All eggs lack sex chromosome
(b) Some of the sperms bear the X-chromosome
(c) All eggs as well as sperms bear the X-chromosome
(d) Some of the eggs bear the X-chromosome
Show Answer & Explanation

Answer: (b) Some of the sperms bear the X-chromosome

Explanation:
1. In insects with XO type sex determination males produce two kinds of sperm, some with X and some without it, whereas all eggs carry X.

Teacher's Note:
1. Sex is decided by the sperm.

Question 15: Loss or gain of a segment of DNA results in
(a) Frame-shift muation
(b) Point mutation
(c) Polyploidy
(d) Chromosomal aberration
Show Answer & Explanation

Answer: (d) Chromosomal aberration

Explanation:
1. Loss or gain of a DNA segment (deletion, duplication) is a chromosomal aberration.

Teacher's Note:
1. Point mutations change a single base.

Question 16: Which one of the following is a physical factor that induce mutation?
(a) Acridines
(b) HNO2
(c) UV-rays
(d) Base analogue
Show Answer & Explanation

Answer: (c) UV-rays

Explanation:
1. UV radiation is a physical mutagen. Acridines, nitrous acid and base analogues are chemical mutagens.

Teacher's Note:
1. Physical mutagens are radiations.

Question 17: If an agouti mouse (CcAa) is crossed with an albino mouse (ccAA), what fraction of the progeny will be albino?
(a) 1/4
(b) 1/2
(c) 3/4
(d) All
Show Answer & Explanation

Answer: (b) 1/2

Explanation:
1. CcAa x ccAA gives CcAA, CcAa, ccAA and ccAa in equal numbers. The cc offspring (2 of 4) are albino, i.e. one half.

Teacher's Note:
1. Albino needs cc regardless of A.

Question 18: Epistasis and dominance are respectively
(a) Intragenic, Intergenic
(b) Non-allelic, Extra-allelic
(c) Non-allelic, Interallelic
(d) Intergenic, Non-allelic
Show Answer & Explanation

Answer: (c) Non-allelic, Interallelic

Explanation:
1. Epistasis is an interaction between non-allelic genes, whereas dominance is an interaction between alleles of the same gene (interallelic).

Teacher's Note:
1. Epistasis = different gene pairs.

Question 19: In a complimentary gene interaction calculate the number of phenotype and genotype produced in a cross AaBb × aaBB
(a) 1 phenotype, 2genotypes
(b) 2 phenotypes, 4 genotypes
(c) 4 phenotypes, 4 genotypes
(d) 2 phenotypes, 2 genotypes
Show Answer & Explanation

Answer: (b) 2 phenotypes, 4 genotypes

Explanation:
1. AaBb x aaBB gives AaBB, AaBb, aaBB, aaBb: 4 genotypes. With complementary genes, A_B_ gives one phenotype and aaB_ another, so 2 phenotypes.

Teacher's Note:
1. Complementary genes give 9 : 7 in F2.

Question 20: If interference is complete or cent percent then the frequency of observed double crossover will be
(a) Equal to expected frequency
(b) Greater than expected frequency
(c) Lesser than expected frequency
(d) Zero
Show Answer & Explanation

Answer: (d) Zero

Explanation:
1. With complete interference there is no double crossing over, so the observed double crossover frequency is zero.

Teacher's Note:
1. Coefficient of coincidence = observed / expected double crossovers.

Question 21: In which of the following disorder a single protein that is a part of the cascade of proteins involved in blood clotting is affected?
(a) Thalassemia
(b) Sickle-cell anaemia
(c) Haemophilia
(d) Phenylketonuria
Show Answer & Explanation

Answer: (c) Haemophilia

Explanation:
1. Haemophilia affects one protein of the blood-clotting cascade (factor VIII in haemophilia A or IX in haemophilia B).

Teacher's Note:
1. Thalassemia and sickle cell involve globin chains.

Question 22: Mark the correct statement (w.r.t. sickle cell-anaemia)
(a) Homozygous individuals for HbS are apparently unaffected
(b) Heterozygous individuals exhibit sickle-cell trait
(c) Heterozygous individuals are affected as well as carrier
(d) Homozygous individuals for HbA show the diseased phenotype
Show Answer & Explanation

Answer: (b) Heterozygous individuals exhibit sickle-cell trait

Explanation:
1. Heterozygotes (HbA HbS) show the sickle-cell trait and are carriers; homozygous HbS individuals are affected.

Teacher's Note:
1. Heterozygotes are generally healthy.

Question 23: Sickle-cell anaemia is caused by the ______ of glutamic acid by valine at the 6th position of the ______ globin chain of haemoglobin.
(a) substitution, beta
(b) deletion, beta
(c) substitution, alpha
(d) duplication, alpha
Show Answer & Explanation

Answer: (a) substitution, beta

Explanation:
1. Sickle-cell anaemia is due to substitution of glutamic acid by valine at the sixth position of the beta globin chain (point mutation GAG to GUG).

Teacher's Note:
1. Beta chain, position 6.

Question 24: A Y-linked gene is responsible for hypertrichosis (long hair on ears). When an affected man marries a normal woman, what percentage of their daughters would be expected to have hairy ears?
(a) 25%
(b) 0%
(c) 50%
(d) 100%
Show Answer & Explanation

Answer: (b) 0%

Explanation:
1. A Y-linked gene passes from father to sons only; daughters do not receive the Y, so 0% of daughters are affected.

Teacher's Note:
1. Holandric genes are never passed to daughters.

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Multiple Choice Questions (MCQs) for NEET Biology Principles of Inheritance and Variation

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