Mendelian Genetics Questions and Answers

Answer: B

When the two alleles of a gene in an individual are different, they are called heterozygous alleles. An individual with Aa is heterozygous for the gene.

Homozygous alleles are identical, and homoduplex and heteroduplex are terms for DNA molecules, not for alleles.

In a heterozygote the two alleles of a gene are different, as in Tt, so the two forms of the gene differ. In a homozygote they are the same, as in TT or tt.

Answer: B

The monohybrid test cross ratio is 1:1.

A test cross crosses an individual of dominant phenotype with the homozygous recessive. For a monohybrid heterozygote, Aa crossed with aa gives half Aa (dominant phenotype) and half aa (recessive phenotype), a 1:1 ratio, which shows the tested parent was heterozygous.

3:1 is the monohybrid F2 ratio, 9:7 a dihybrid F2 ratio modified by complementary gene action, and 1:7 is not a standard ratio. The dihybrid test cross ratio is 1:1:1:1.

Answer: A

This is called Incomplete dominance.

In incomplete dominance, neither allele fully masks the other, so the heterozygote shows an intermediate phenotype. The classic example is the four o'clock plant (Mirabilis jalapa): red crossed with white gives pink, and the F2 segregates 1 red : 2 pink : 1 white.

In complete dominance the heterozygote looks like the dominant homozygote. In overdominance it exceeds both homozygotes, the basis of one theory of heterosis. Pseudo-dominance is the appearance of a recessive trait when the dominant allele is lost by deletion.

Answer: D

The F1 progeny is Heterozygous and Homogeneous.

When two pure (homozygous) lines are crossed, every F1 plant receives one set of alleles from each parent, so each plant is heterozygous at all loci where the parents differ. And because every plant gets exactly the same combination, the F1 population is genetically uniform, or homogeneous. This is why an F1 hybrid crop is uniform.

The F2 is heterozygous and heterogeneous because the genes segregate, while a pure line is homozygous and homogeneous. A homozygous but heterogeneous population is a mixture of different pure lines, such as a landrace.

Answer: C

The F2 has 81 different genotypes.

Each heterozygous locus (Aa) segregates into three genotypes in the F2: AA, Aa and aa. Loci assort independently, so the counts multiply:

\[ \begin{aligned} \text{genotypes} &= 3^{n} \\ &= 3^{4} = 81 \end{aligned} \]

The distractors come from the other standard formulas: \(2^{4} = 16\) is the number of gamete types and of phenotypes with complete dominance, \(4^{4}\) gives the number of gamete combinations, and \(3^{3} = 27\) would apply to only three loci. 64 is \(4^{3}\), which fits none of these.

Answer: A

The example is flower color in Mirabilis jalapa.

In the four o'clock plant, a cross between red-flowered (RR) and white-flowered (rr) plants gives pink F1 plants (Rr), a blend of the two parents. In the F2 the colours appear in the ratio 1 red : 2 pink : 1 white, so phenotype and genotype ratios coincide. That intermediate heterozygote is the mark of incomplete dominance.

Seed shape in pea shows complete dominance of round over wrinkled. Blood group AB shows codominance, where both alleles are fully expressed, not blended. Coat colour in rabbit is the standard example of multiple alleles.

Answer: D

The sex-linked trait is Colour blindness.

Red-green colour blindness is caused by a recessive allele on the X chromosome. A man has only one X, so a single copy of the allele makes him colour blind, while a woman needs two copies. That is why the condition is far more common in men and is passed from a carrier mother to half her sons.

Thalassemia, sickle cell anaemia and albinism are all controlled by genes on autosomes, not on the sex chromosomes, so they affect men and women equally and are not sex-linked.

Answer: A

The probability is 1/64.

The three genes assort independently, so the probability for each gene can be worked out separately and multiplied. In \(Aa \times Aa\) the chance of \(aa\) is 1/4, and likewise for \(bb\) and \(cc\):

\[ \begin{aligned} P(aabbcc) &= \tfrac{1}{4} \times \tfrac{1}{4} \times \tfrac{1}{4} \\ &= \tfrac{1}{64} \end{aligned} \]

Equivalently, each parent makes \(2^{3} = 8\) kinds of gamete, giving \(8 \times 8 = 64\) combinations, of which only one (abc from each parent) is fully recessive.

1/16 is the answer for a dihybrid cross, and 1/32 and 1/128 come from mixing up the powers of 2 and 4.