Principles of Inheritance and Variation NEET PYQReal NEET & AIPMT 2005–2021 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
Every Principles of Inheritance and Variation question from the NEET & AIPMT papers on this site (2005–2021), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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Principles of Inheritance and Variation previous year questions
84 questions. Tap an option to check your answer.
Q1NEET 2021
The production of gametes by the parents, formation of zygotes, the F₁ and F₂ plants, can be understood from a diagram called :
Not quite — the answer is C.
The Punnett square, devised by Reginald C. Punnett, is the checkerboard diagram that shows gametes of both parents and the zygote combinations of the F₁ and F₂ generations. The other names are not real genetic terms.
Q2NEET 2021
In a cross between a male and female, both heterozygous for sickle cell anaemia gene, what percentage of the progeny will be diseased ?
Not quite — the answer is C.
Sickle cell anaemia is autosomal recessive, so Hbᴬ Hbˢ × Hbᴬ Hbˢ gives 1 Hbᴬ Hbᴬ : 2 Hbᴬ Hbˢ : 1 Hbˢ Hbˢ. Only the Hbˢ Hbˢ quarter is diseased, i.e. 25%; the 50% heterozygotes are carriers, not patients.
Q3NEET 2020 · Q6
Identify the wrong statement with reference to the gene 'I' that controls ABO blood groups.
Not quite — the answer is C.
Iᴬ and Iᴮ are codominant: each adds a different sugar to the plasma membrane, so a person with both shows both A and B antigens (blood group AB). The other statements are correct: gene I has three alleles, a diploid person carries any two of them, and allele i adds no sugar.
Q4NEET 2020 · Q8
Select the correct match.
Not quite — the answer is C.
Sickle cell anaemia is an autosomal recessive disorder due to a mutation in the beta-globin gene on chromosome 11. Haemophilia is X-linked recessive, phenylketonuria is autosomal recessive and thalassemia is autosomal recessive, so those pairs are wrong.
Q5NEET 2020 · Q61
How many true breeding pea plant varieties did Mendel select as pairs, which were similar except in one character with contrasting traits?
Not quite — the answer is C.
Mendel chose 14 true-breeding pea varieties as 7 pairs, each pair differing in one character with contrasting traits. The trap is to answer 7, the number of characters, or 2, the number of traits per character.
Q6NEET 2020 · Q83
Experimental verification of the chromosomal theory of inheritance was done by:
Not quite — the answer is D.
Thomas Hunt Morgan's work on Drosophila gave the experimental verification of the chromosomal theory of inheritance. Sutton and Boveri proposed the theory, and Mendel worked before chromosomes were linked to heredity.
Q7NEET 2019 · Q121
Select the incorrect statement.
Not quite — the answer is C.
Birds show ZW–ZZ sex determination: females (ZW) are heterogametic, so the sex of the chick depends on the egg, not the sperm. Male Drosophila are XY, male grasshoppers are XO (half the sperms lack an X), and the human Y is much shorter than the X.
Q8NEET 2019 · Q129
What map unit (Centimorgan) is adopted in the construction of genetic maps ?
Not quite — the answer is C.
One centimorgan (map unit) is the distance between genes on a chromosome that gives 1% recombination (crossing over). 50% recombination is the maximum and indicates unlinked genes.
Q9NEET 2019 · Q147
In Antirrhinum (Snapdragon), a red flower was crossed with a white flower and in F₁ generation, pink flowers were obtained. When pink flowers were selfed, the F₂ generation showed white, red and pink flowers. Choose the incorrect statement from the following :
Not quite — the answer is D.
In incomplete dominance the alleles still segregate during gamete formation, which is why the F₂ shows a 1 red : 2 pink : 1 white ratio; so saying the law of segregation does not apply is incorrect. Only the principle of dominance fails here, because the heterozygote is intermediate.
Q10NEET 2019 · Q165
The frequency of recombination between gene pairs on the same chromosome as a measure of the distance between genes was explained by :
Not quite — the answer is C.
Alfred Sturtevant, Morgan's student, used the frequency of recombination between gene pairs on the same chromosome as a measure of distance and mapped genes. Morgan discovered linkage and coined the term recombination but did not build the map.
Q11NEET 2019 · Q176
What is the genetic disorder in which an individual has an overall masculine development, gynaecomastia, and is sterile ?
Not quite — the answer is B.
Klinefelter's syndrome (47, XXY) gives overall masculine development with feminine features such as gynaecomastia, and the individuals are sterile. Turner's syndrome (45, XO) affects females.
Q12NEET 2018 · Q104
Which of the following pairs is wrongly matched?
Not quite — the answer is C.
Starch synthesis in pea is controlled by one gene with two alleles (B and b); it is NCERT's example of incomplete dominance at the level of starch grain size and of pleiotropy, not of multiple alleles. Grasshopper has XO sex determination, the Iᴬ and Iᴮ alleles of ABO are co-dominant and Morgan worked out linkage.
Q13NEET 2018 · Q105
Select the correct statement:
Not quite — the answer is B.
The Punnett square is named after Reginald C. Punnett, a British geneticist. Spliceosomes act in splicing of hnRNA, not translation; Morgan coined the term linkage; and transduction was discovered by Zinder and Lederberg, while Altman is known for ribozymes.
Q14NEET 2018 · Q159
A woman has an X-linked condition on one of her X chromosomes. This chromosome can be inherited by
Not quite — the answer is D.
A mother passes one of her two X chromosomes to every child, sons and daughters alike, so either sex can receive the affected X. It is the father's X that goes only to daughters, which is the trap.
Q15NEET 2018 · Q180
Which of the following characteristics represent 'Inheritance of blood groups' in humans? a. Dominance b. Co-dominance c. Multiple allele d. Incomplete dominance e. Polygenic inheritance
Not quite — the answer is B.
ABO blood groups are controlled by one gene I with three alleles (multiple alleles); Iᴬ and Iᴮ are each dominant over i (dominance), and when together both are expressed (co-dominance). Blood group is not a polygenic trait and shows no incomplete dominance.
Q16NEET 2017 · Q93
Thalassemia and sickle cell anemia are caused due to a problem in globin molecule synthesis. Select the correct statement.
Not quite — the answer is C.
Thalassemia is a quantitative problem in which too few globin chains are made. Sickle cell anaemia is a qualitative problem, an abnormal β-globin with valine in place of glutamic acid, so calling it quantitative is wrong.
Q17NEET 2017 · Q104
Which one from those given below is the period for Mendel's hybridization experiments?
Not quite — the answer is A.
Mendel carried out his hybridisation experiments on garden pea for seven years, from 1856 to 1863, and published them in 1865. The other ranges do not match this period.
Q18NEET 2017 · Q118
The genotypes of a Husband and Wife are IᴬIᴮ and Iᴬi. Among the blood types of their children, how many different genotypes and phenotypes are possible?
Not quite — the answer is C.
Crossing IᴬIᴮ × Iᴬi gives IᴬIᴬ, Iᴬi, IᴬIᴮ and Iᴮi, which is 4 genotypes. IᴬIᴬ and Iᴬi are both blood group A, so the phenotypes are A, AB and B, which is 3. Group O cannot appear because no child can be ii.
Q19NEET 2017 (Gujarati paper) · Q58
The recessive genes located on X-chromosome in humans are always:
Not quite — the answer is B.
Males are hemizygous (XY), so a single recessive allele on their only X chromosome is always expressed, as in haemophilia and colour blindness. Females express it only when homozygous, and most such genes are neither lethal nor sub-lethal.
Q20NEET 2017 (Gujarati paper) · Q64
Select the correct option:
Not quite — the answer is D.
Klinefelter's syndrome (47, XXY) results from an extra X chromosome and gives a sterile male. Phenylketonuria is autosomal recessive, Down's syndrome is trisomy of chromosome 21 (not triploidy), and Turner's syndrome is a monosomy (45, XO).
Q21NEET 2017 (Gujarati paper) · Q118
Select the wrong statement:
Not quite — the answer is B.
Linkage and recombination were discovered by T.H. Morgan through his Drosophila experiments, not by Sutton. Sutton (with Boveri) proposed the chromosomal theory of inheritance, and de Vries, Correns and von Tschermak rediscovered Mendel's work in 1900.
Q22NEET 2017 (Gujarati paper) · Q122
Which one of the following is not true for the experiments of Mendel on pea?
Not quite — the answer is C.
Mendel made his crosses by controlled artificial pollination (emasculation and bagging), not by natural open pollination. He did choose seven characters with two contrasting forms, start from true-breeding lines and use large samples, which gave his data statistical reliability.
Q23NEET 2016 Phase 2 · Q121
The mechanism that causes a gene to move from one linkage group to another is called
Not quite — the answer is C.
In translocation a chromosome segment breaks off and joins a non-homologous chromosome, so its genes shift to a different linkage group. Crossing over exchanges segments between homologous chromosomes of the same linkage group, and inversion and duplication change order or copy number within a chromosome.
Q24NEET 2016 Phase 2 · Q123
A true breeding plant is
Not quite — the answer is C.
A true-breeding line has been self-pollinated for generations, making it nearly homozygous, so it gives offspring with the same trait. It may be homozygous dominant or recessive, and cross-pollination among unrelated plants increases heterozygosity instead.
Q25NEET 2016 Phase 2 · Q165
If a colour-blind man marries a woman who is homozygous for normal colour vision, the probability of their son being colour-blind is
Not quite — the answer is A.
Colour blindness is X-linked recessive, and a son gets his only X chromosome from his mother. Here the mother is XX with two normal alleles and the father passes only his Y to sons, so every son is normal and the probability is 0; the daughters will be carriers.
Q26NEET 2016 Phase 1 · Q96
Which of the following most appropriately describes haemophilia?
Not quite — the answer is B.
Haemophilia is caused by a recessive allele on the X chromosome, so it is a sex-linked (X-linked) recessive disorder, seen mostly in males. 'Recessive gene disorder' is true but less precise; it is not a chromosomal or dominant disorder.
Q27NEET 2016 Phase 1 · Q119
A tall true breeding garden pea plant is crossed with a dwarf true breeding garden pea plant. When the F₁ plants were selfed the resulting genotypes were in the ratio of:
Not quite — the answer is A.
TT × tt gives Tt in F₁; selfing Tt gives TT : Tt : tt in a 1 : 2 : 1 ratio, i.e. tall homozygous : tall heterozygous : dwarf. 3 : 1 tall : dwarf is the phenotypic ratio, not the genotypic one asked for.
Q28NEET 2016 Phase 1 · Q126
Match the terms in Column I with their description in Column II and choose the correct option: Column I: (a) Dominance (b) Codominance (c) Pleiotropy (d) Polygenic inheritance Column II: (i) Many genes govern a single character (ii) In a heterozygous organism only one allele expresses itself (iii) In a heterozygous organism both alleles express themselves fully (iv) A single gene influences many characters
Not quite — the answer is B.
Dominance means only one allele is expressed in the heterozygote; codominance means both alleles are fully expressed (e.g. ABO blood group); pleiotropy is one gene affecting several traits (e.g. phenylketonuria); polygenic inheritance is many genes controlling one trait (e.g. skin colour).
Q29NEET 2016 Phase 1 · Q128
Pick out the correct statements: (a) Haemophilia is a sex-linked recessive disease. (b) Down's syndrome is due to aneuploidy. (c) Phenylketonuria is an autosomal recessive gene disorder. (d) Sickle cell anaemia is an X - linked recessive gene disorder.
Not quite — the answer is D.
Haemophilia is X-linked recessive, Down's syndrome results from trisomy 21 (aneuploidy) and phenylketonuria is autosomal recessive. Sickle-cell anaemia is autosomal recessive (HbS gene on chromosome 11), not X-linked, so every combination containing it is wrong.
Q30NEET 2016 Phase 1 · Q170
In a testcross involving F₁ dihybrid flies, more parental-type offspring were produced than the recombinant-type offspring. This indicates:
Not quite — the answer is C.
Independent assortment would give equal parental and recombinant classes (1:1:1:1) in a dihybrid testcross. An excess of parental types means the genes tend to be inherited together, i.e. they are linked on the same chromosome, with recombinants coming only from crossing over.
Q31NEET 2016 Phase 1 · Q179
A cell at telophase stage is observed by a student in a plant brought from the field. He tells his teacher that this cell is not like other cells at telophase stage. There is no formation of cell plate and thus the cell is containing more number of chromosomes as compared to other dividing cells. This would result in:
Not quite — the answer is B.
If chromosomes separate but no cell plate forms (failure of cytokinesis), both sets stay in one cell and the chromosome number doubles, i.e. polyploidy. Aneuploidy is gain or loss of single chromosomes, and polyteny is repeated DNA replication without separation of chromatids.
Q32AIPMT 2015
Alleles are:
Not quite — the answer is D.
Alleles are alternative forms of the same gene at one locus, differing slightly in their DNA sequence. Heterozygotes and true-breeding homozygotes are kinds of individuals, and a phenotype is the visible trait, not the gene form.
Q33AIPMT 2015
The movement of a gene from one linkage group to another is called:
Not quite — the answer is D.
Translocation moves a chromosome segment, with its genes, to a non-homologous chromosome, which is a different linkage group. Crossing over only exchanges segments between homologous chromosomes of the same linkage group.
Q34AIPMT 2015
Multiple alleles are present:
Not quite — the answer is D.
Multiple alleles are more than two alternative forms of one gene, so they all occupy the same locus (e.g. Iᴬ, Iᴮ and i of the ABO gene). Genes at different loci or on different chromosomes are different genes, not alleles.
Q35AIPMT 2015
A man with blood group 'A' marries a woman with blood group 'B'. What are all the possible blood groups of their offsprings?
Not quite — the answer is D.
If both parents are heterozygous (Iᴬi × Iᴮi), the children can be IᴬIᴮ (group AB), Iᴬi (group A), Iᴮi (group B) or ii (group O). So all four groups are possible; 'A, B and AB only' would apply only if one parent were homozygous.
Q36AIPMT 2015
How many pairs of contrasting characters in pea plants were studied by Mendel in his experiments?
Not quite — the answer is A.
Mendel studied seven pairs of contrasting characters in pea, such as tall/dwarf stem, round/wrinkled seed and yellow/green seed.
Q37AIPMT 2014
Fruit colour in squash is an example of:
Not quite — the answer is A.
In summer squash the dominant allele for white fruit colour masks the expression of the gene for yellow versus green, giving a 12 : 3 : 1 ratio in F₂, which is dominant epistasis. Recessive epistasis gives 9 : 3 : 4 and complementary genes give 9 : 7, so those do not fit.
Q38AIPMT 2014
A man whose father was colour blind marries a woman who had a colour blind mother and normal father. What percentage of male children of this couple will be colour blind?
Not quite — the answer is B.
The woman's colour-blind mother must give her Xᶜ and her normal father gives a normal X, so she is a carrier (XXᶜ). Sons receive their X only from the mother, so half the sons get Xᶜ and are colour blind: 50%. The man's father being colour blind does not matter, because a father never passes his X to his sons and the man himself is normal.
Q39AIPMT 2014
A human female with Turner's syndrome:
Not quite — the answer is D.
Turner's syndrome results from absence of one X chromosome, giving 44 + XO = 45 chromosomes. Such females are sterile with rudimentary ovaries, so they cannot bear children, and they do not have an extra X (that is Klinefelter's XXY in males) or male characters.
Q40NEET 2013
Which of the following statements is not true of two genes that show 50% recombination frequency?
Not quite — the answer is B.
A recombination frequency of 50% is the maximum and is what unlinked genes show, so such genes behave as if assorting independently, whether on different chromosomes or far apart on one chromosome with multiple crossovers. Tightly linked genes show very low recombination, so calling them tightly linked is the untrue statement.
Q41NEET 2013
If two persons with 'AB' blood group marry and have sufficiently large number of children, these children could be classified as 'A' blood group : 'AB' blood group : 'B' blood group in 1 : 2 : 1 ratio. Modern technique of protein electrophoresis reveals presence of both 'A' and 'B' type proteins in 'AB' blood group individuals. This is an example of
Not quite — the answer is A.
In AB individuals both the IA and IB alleles are fully expressed, producing both A and B proteins, which is codominance. In incomplete dominance the heterozygote would have an intermediate phenotype instead of showing both parental products.
Q42NEET 2013
The incorrect statement with regard to haemophilia is
Not quite — the answer is C.
Haemophilia is an X-linked recessive disorder in which a single clotting factor protein is affected. Calling it a dominant disease is the incorrect statement.
Q43NEET 2013
If both parents are carriers for thalassemia, which is an autosomal recessive disorder, what are the chances of pregnancy resulting in an affected child?
Not quite — the answer is C.
Two carriers (Tt × Tt) give TT : Tt : tt in a 1 : 2 : 1 ratio, so the chance of an affected (tt) child is 1/4, i.e. 25%. 50% is the chance of a carrier child.
Q44AIPMT 2012 Mains
Which one of the following is a wrong statement regarding mutations ?
Not quite — the answer is D.
A change in a single base pair is itself a mutation (a point mutation, as in sickle-cell anaemia), so that statement is wrong. Insertions and deletions do shift the reading frame, cancer cells commonly show chromosomal aberrations, and UV and gamma rays are mutagens.
Q45AIPMT 2012 Mains
A test cross is carried out to
Not quite — the answer is A.
A test cross crosses a dominant-phenotype plant (such as one from the F₂) with the homozygous recessive parent; the offspring ratio reveals whether that plant is homozygous or heterozygous. It is not a tool for testing linkage or counting alleles.
Q46AIPMT 2012 Prelims
A normal-visioned man whose father was colour-blind, marries a woman whose father was also colour-blind. They have their first child as a daughter. What are the chances that this child would be colour-blind ?
Not quite — the answer is C.
The man has normal vision, so his X chromosome carries the normal allele; every daughter receives this X from him. The woman, whose father was colour-blind, is a carrier, but the daughter can be at most a carrier, so her chance of being colour-blind is zero.
Q47AIPMT 2011 Mains
Which one of the following conditions of the zygotic cell would lead to the birth of a normal human female child?
Not quite — the answer is C.
A normal human female has 44 autosomes plus two X chromosomes (44 + XX). A single X gives Turner's syndrome, X with Y gives a male, and a zygote with only a Y chromosome is not viable.
Q48AIPMT 2011 Mains
Test cross in plants or in Drosophila involves crossing
Not quite — the answer is A.
A test cross crosses an organism showing the dominant phenotype, typically the F₁ hybrid, with the recessive parent; for two genes that is the double recessive genotype. Crossing two F₁ hybrids is a selfing or F₂ cross, not a test cross.
Q49AIPMT 2011 Prelims
Which one of the following conditions correctly describes the manner of determining the sex in the given example?
Not quite — the answer is C.
In grasshoppers, males have only one X chromosome (XO) and females have XX, so the XO condition determines the male sex. In Drosophila XX is female, in birds ZZ is male, and in humans sex is decided by the presence or absence of the Y chromosome.
Q50AIPMT 2010 Mains · Q33
In Antirrhinum two plants with pink flowers were hybridized. The F₁ plants produced red, pink and white flowers in the proportion of 1 red, 2 pink and 1 white. What could be the genotype of the two plants used for hybridization? Red flower colour is determined by RR, and white by rr genes.
Not quite — the answer is C.
Flower colour in Antirrhinum (dog flower) shows incomplete dominance, so the heterozygote Rr is pink. Crossing Rr × Rr gives RR : Rr : rr in a 1 : 2 : 1 ratio, i.e. red : pink : white. RR or rr parents would breed true for red or white.
Q51AIPMT 2010 Mains · Q36
A cross in which an organism showing a dominant phenotype is crossed with the recessive parent in order to know its genotype is called :
Not quite — the answer is C.
Crossing a dominant-phenotype individual of unknown genotype with a homozygous recessive is a test cross: all dominant offspring means the individual was homozygous, and a 1 : 1 ratio means it was heterozygous. A back cross is any cross of F₁ with either parent, so only the cross with the recessive parent is the test cross.
Q52AIPMT 2010 Mains · Q42
ABO blood grouping is controlled by gene I which has three alleles and show co-dominance. There are six genotypes. How many phenotypes in all are possible?
Not quite — the answer is C.
The three alleles Iᴬ, Iᴮ and i give six genotypes, but Iᴬ and Iᴮ are dominant over i and co-dominant with each other, so only four phenotypes appear: A, B, AB and O. Six would be correct only if every genotype looked different.
Q53AIPMT 2010 Mains · Q51
The fruit fly Drosophila melanogaster was found to be very suitable for experimental verification of chromosomal theory of inheritance by Morgan and his colleagues because :
Not quite — the answer is D.
Drosophila is useful for genetics because it completes its life cycle in about two weeks, a single mating gives a large number of progeny, and the sexes are easy to tell apart. The female is larger than the male, not smaller, and flies reproduce sexually, not parthenogenetically.
Q54AIPMT 2010 Prelims · Q119
Which one of the following cannot be explained on the basis of Mendel’s Law of Dominance?
Not quite — the answer is D.
The return of both parental traits unchanged in the F₂, without blending, is explained by the Law of Segregation (purity of gametes), not by the Law of Dominance. Paired factors and one dominant factor masking a recessive one are the ideas behind dominance.
Q55AIPMT 2010 Prelims · Q120
The genotype of a plant showing the dominant phenotype can be determined by
Not quite — the answer is B.
In a test cross, the dominant-phenotype plant is crossed with a homozygous recessive. All dominant offspring mean it is homozygous, and a 1:1 ratio means it is heterozygous. A back cross can be made to either parent, so it does not necessarily reveal the genotype.
Q56AIPMT 2010 Prelims · Q124
ABO blood groups in humans are controlled by the gene I. It has three alleles–Iᴬ, Iᴮ and i. Since there are three different alleles, six different genotypes are possible. How many phenotypes can occur?
Not quite — the answer is D.
The six genotypes give four blood groups: A (IᴬIᴬ, Iᴬi), B (IᴮIᴮ, Iᴮi), AB (IᴬIᴮ, codominance) and O (ii). The answer 'three' misses the AB group, where both alleles are expressed.
Q57AIPMT 2010 Prelims · Q125
Select the correct statement from the ones given below with respect to dihybrid cross
Not quite — the answer is A.
Morgan found that genes lying close together (tightly linked) rarely separate by crossing over, so they show few recombinants, as with the white-eye and yellow-body genes in Drosophila. Genes far apart or loosely linked show more recombination, not less.
Q58AIPMT 2010 Prelims · Q127
Which one of the following symbols and its representation, used in human pedigree analysis is correct?
Not quite — the answer is B.
In pedigree charts, a double line joining a male (square) and a female (circle) shows a consanguineous mating, that is, mating between relatives. A shaded diamond is an affected person of unspecified sex, a plain circle is an unaffected female, and a plain square is an unaffected male.
Q59AIPMT 2009 Prelims · Q77
Sickle cell anemia is:
Not quite — the answer is B.
Sickle cell anaemia results from a single base substitution (GAG to GUG) in the β-globin gene, a point mutation, so glutamic acid at the sixth position is replaced by valine, not the other way round. The trait is autosomal recessive and mature RBCs have no nucleus.
Q60AIPMT 2009 Prelims · Q123
Select the incorrect statement from the following:
Not quite — the answer is C.
Baldness is a sex-influenced trait (expressed more often in males because of hormones), not a sex-limited trait, which is expressed in only one sex. The other statements are correct: galactosemia is an inborn error, genetic drift is strong in small populations, and linked genes do not assort independently.
Q61AIPMT 2009 Prelims · Q148
Point mutation involves:
Not quite — the answer is A.
A point mutation is a change in a single base pair of DNA, as in sickle cell anaemia. Duplication and larger deletions or insertions alter segments of chromosomes or DNA rather than a single base.
Q62AIPMT 2008 Prelims · Q120
Haploids are more suitable for mutation studies than the diploids. This is because
Not quite — the answer is A.
A haploid has only one copy of each gene, so a recessive mutation cannot be masked by a dominant allele and shows up directly in the phenotype. Mutagen penetration does not depend on ploidy, so that option gives a wrong reason.
Q63AIPMT 2008 Prelims · Q122
Which one of the following conditions in humans is correctly matched with its chromosomal abnormality/ linkage?
Not quite — the answer is B.
Klinefelter syndrome results from an extra X chromosome in a male, giving 44 autosomes + XXY (47 chromosomes). Down syndrome is trisomy 21 (44 + XO is Turner syndrome), colour blindness is X-linked, and erythroblastosis foetalis comes from Rh incompatibility, not sex linkage.
Q64AIPMT 2007 Prelims · Q138
In the hexaploid wheat, the haploid (n) and basic (x) numbers of chromosomes are
Not quite — the answer is C.
Bread wheat is hexaploid with 2n = 6x = 42. Its gametes carry n = 21 chromosomes, and the basic set is x = 42/6 = 7. Options that equate n and x ignore that n here contains three basic sets.
Q65AIPMT 2007 Prelims · Q157
Two genes R and Y are located very close on the chromosomal linkage map of maize plant. When RRYY and rryy genotypes are hybridized, the F2 segregation will show
Not quite — the answer is C.
Genes lying very close together are tightly linked and rarely separated by crossing over, so the F2 shows many more parental combinations than recombinants. The 9:3:3:1 ratio holds only for independently assorting genes.
Q66AIPMT 2007 Prelims · Q161
Inheritance of skin colour in humans is an example of
Not quite — the answer is B.
Human skin colour is controlled by several genes with additive effects, giving continuous variation; this is polygenic inheritance. Codominance involves two alleles of one gene both being expressed, as in ABO blood groups.
Q67AIPMT 2007 Prelims · Q178
In pea plants, yellow seeds are dominant to green. If a heterozygous yellow seeded plant is crossed with a green seeded plant, what ratio of yellow and green seeded plants would you expect in F1 generation?
Not quite — the answer is D.
Yy x yy gives half Yy (yellow) and half yy (green), a 1:1 or 50:50 ratio. A 3:1 ratio would need both parents to be heterozygous.
Q68AIPMT 2007 Prelims · Q180
A human male produces sperms with the genotypes AB, Ab, aB, and ab pertaining to two diallelic characters in equal proportions. What is the corresponding genotype of this person?
Not quite — the answer is D.
Four gamete types in equal proportions are produced only by a double heterozygote (AaBb) with independent assortment. AaBB or AABb would give only two gamete types, and AABB only one.
Q69AIPMT 2006 Prelims · Q120
Which one of the following is the most suitable, medium for culture of Drosophila melanogaster ?
Not quite — the answer is C.
Drosophila feeds on yeasts in fermenting fruit, so ripe (overripe) banana is the classic simple culture medium. Agar alone has no food value for the flies.
Q70AIPMT 2006 Prelims · Q123
Phenotype of an organism is the result of :
Not quite — the answer is D.
The observable phenotype is produced by the genotype acting within a given environment. Mutations, nutrition or sexual dimorphism each affect only a part of it.
Q71AIPMT 2006 Prelims · Q132
How many different kinds of gametes will be produced by a plant having the genotype AABbCC ?
Not quite — the answer is D.
Only the B gene is heterozygous, so the plant makes 2^1 = 2 kinds of gametes: ABC and AbC. AA and CC each give only one kind of allele.
Q72AIPMT 2006 Prelims · Q162
Which one of the following is an example of polygenic inheritance ?
Not quite — the answer is D.
Human skin colour is controlled by several genes with additive effects, giving continuous variation; this is polygenic inheritance. Flower colour in Mirabilis jalapa shows incomplete dominance of a single gene.
Q73AIPMT 2006 Prelims · Q171
Test cross involves :
Not quite — the answer is C.
In a test cross the F₁ (or any dominant-phenotype) plant is crossed with the homozygous recessive parent to find its genotype. Crossing two F₁ hybrids is selfing that gives the F₂.
Q74AIPMT 2006 Prelims · Q175
If a colourblind woman marries a normal visioned man, their sons will be :
Not quite — the answer is D.
A colourblind woman is XᶜXᶜ, so every son gets an X carrying the colourblind allele from her and a Y from the father; all sons are colourblind. The daughters would be carriers with normal vision.
Q75AIPMT 2006 Prelims · Q197
Sickle cell anaemia has not been eliminated from the African population because :
Not quite — the answer is C.
Heterozygous carriers (HbA HbS) are more resistant to falciparum malaria, so the allele is kept in high frequency where malaria is common. Simply being recessive does not explain why the allele persists at such a high frequency.
Q76AIPMT 2006 Prelims · Q199
Both sickle cell anaemia and Huntington's chorea are :
Not quite — the answer is B.
Both sickle cell anaemia and Huntington's chorea are genetic (inherited) disorders, present from birth in the genotype. They are not caused by microbes or pollutants.
Q77AIPMT 2005 Prelims · Q112
A man and a woman, who do not show any apparent signs of a certain inherited disease, have seven children (2 daughters and 5 sons). Three of the sons suffer from the given disease but none of the daughters are affected. Which of the following mode of inheritance do you suggest for this disease ?
Not quite — the answer is D.
Unaffected parents with affected sons only fits an X-linked recessive trait: the mother is a carrier and passes the mutant X to some sons, while daughters receive the father's normal X and are not affected. A dominant trait would show up in at least one of the parents.
Q78AIPMT 2005 Prelims · Q125
G-6-P dehydrogenase deficiency is associated with haemolysis of :
Not quite — the answer is B.
Red blood cells rely on glucose-6-phosphate dehydrogenase to make NADPH, which protects them from oxidative damage. In the X-linked deficiency, oxidant drugs or fava beans cause RBCs to break down (haemolytic anaemia). White cells and platelets are not the cells that lyse.
Q79AIPMT 2005 Prelims · Q154
A woman with normal vision, but whose father was colour blind, marries a colourblind man. Suppose that the fourth child of this couple was a boy. This boy :
Not quite — the answer is D.
The woman got her colour-blind father's X, so she is a carrier (XᶜX). A son always gets his X from his mother, so each son has a 50% chance of colour blindness, whatever the birth order. Males carry only one X, so a son cannot be heterozygous.
Q80AIPMT 2005 Prelims · Q159
Which of the following is not a hereditary disease ?
Not quite — the answer is A.
Cretinism is caused by thyroid hormone deficiency in childhood (often iodine deficiency), a hormonal disorder rather than an inherited gene defect. Cystic fibrosis, thalassaemia and haemophilia are all Mendelian genetic disorders.
Q81AIPMT 2005 Prelims · Q184
Haemophilia is more commonly seen in human males than in human females because :
Not quite — the answer is C.
Haemophilia is X-linked recessive. Males have one X, so one mutant allele causes the disease, while females need two copies and are usually carriers. A Y-linked trait would never appear in females at all.
Q82AIPMT 2005 Prelims · Q185
A woman with 47 chromosomes due to three copies of chromosome 21 is characterized by :
Not quite — the answer is A.
Trisomy 21 (47 chromosomes) is Down syndrome. Turner syndrome is 45,XO, super female is XXX (trisomy of the X), and triploidy would mean three full sets (69 chromosomes).
Q83AIPMT 2005 Prelims · Q186
In order to find out the different types of gametes produced by a pea plant having the genotype AaBb, it should be crossed to a plant with the genotype
Not quite — the answer is D.
A test cross with the double recessive aabb works because that parent gives only ab gametes, so each offspring's phenotype shows exactly which gamete came from the AaBb plant. Dominant alleles in the tester would hide the recessive gametes.
Q84AIPMT 2005 Prelims · Q196
The salivary gland chromosomes in the dipteran larvae, are useful in gene mapping because :
Not quite — the answer is D.
Polytene chromosomes form by repeated DNA replication without cell division (endoreduplication), so many chromatids lie side by side. This gives clear band patterns that can be linked to particular genes. Their large size is a result of endoreduplication, and the bands are what make mapping possible.
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