Genetics Questions and Answers

Answer: C

The pattern is best explained by Mitochondrial inheritance.

When the mother is mutant, every offspring of both sexes is mutant; when only the father is mutant, every offspring is wild type. The trait follows the mother alone, which is the signature of maternal (cytoplasmic) inheritance: mitochondria pass to the offspring through the egg, while the sperm contributes essentially none.

A recessive or autosomal nuclear gene would behave the same in reciprocal crosses. An X-linked (sex-linked) gene would give different results in sons and daughters, not the same result in all eight offspring of each cross.

Answer: C

The likely explanation is that a suppressor for the tall allele is present in the dwarf plant.

In Mendel's cross every F1 plant receives one tall allele, which is dominant, so all are tall. Getting only short plants from the same cross means the tall allele is present but not expressed. A dominant suppressor at another locus, carried by the dwarf parent and passed to every F1 plant, would mask the tall allele in all of them. This is gene interaction (epistasis) rather than a failure of dominance.

Heterozygous tall parents would give about half tall and half dwarf offspring, not all short. An enhancer of the tall allele would, if anything, make the plants taller. The dwarf parent being homozygous is the normal situation in Mendel's cross and predicts all tall F1, not all short.

Answer: C

The cross gives 27 genotypes and 8 phenotypes.

With independent assortment, count each gene pair separately and multiply:

• Aa × Aa: 3 genotypes (AA, Aa, aa), 2 phenotypes.
• BB × BB: 1 genotype (BB), 1 phenotype.
• Cc × Cc: 3 genotypes, 2 phenotypes.
• Dd × Dd: 3 genotypes, 2 phenotypes.

\[ \begin{aligned} \text{genotypes} &= 3 \times 1 \times 3 \times 3 = 27 \\ \text{phenotypes} &= 2 \times 1 \times 2 \times 2 = 8 \end{aligned} \]

8 and 4 count only phenotypes and miss one gene; 12 and 4 and 14 and 8 do not come from multiplying the per-gene counts. The common slip is to treat BB × BB as a segregating pair, which would give 81 genotypes and 16 phenotypes.

Answer: B

The frequency of AABB offspring is 1/16.

The two genes assort independently, so treat each separately. From Aa x Aa the chance of AA is 1/4, and from Bb x Bb the chance of BB is 1/4:

\[ \begin{aligned} P(\mathrm{AABB}) &= \frac{1}{4} \times \frac{1}{4} \\ &= \frac{1}{16} \end{aligned} \]

This is the one homozygous dominant square in the 16-square dihybrid Punnett square. 3/16 is the share of one of the single-dominant phenotype classes (A_bb or aaB_) in the 9:3:3:1 ratio, 1/4 is the probability for one gene alone, and 1/64 would apply to a trihybrid cross.

Answer: C

The progeny is called a pureline.

Johannsen (1903) used the term pure line for the descendants of a single homozygous self-fertilised plant. Because the parent is homozygous at all loci and self-pollination keeps it so, every offspring is genetically identical to it, and selection within a pure line has no effect.

An inbred line comes from several generations of inbreeding of a population and need not start from one homozygous plant. Isogenic and near-isogenic lines are lines made to differ at one or a few chosen genes, usually by backcrossing.

Answer: C

The expected ratio is 1 : 1 : 1 : 1.

This is a dihybrid test cross. The double heterozygote (PpHh) makes four kinds of gamete in equal numbers, PH, Ph, pH and ph, and the double recessive (pphh) makes only ph. Each offspring phenotype therefore reflects one gamete of the heterozygote:

\[ \text{PpHh} \times \text{pphh} \to 1 : 1 : 1 : 1 \]

9 : 3 : 3 : 1 is the ratio from crossing two double heterozygotes. 9 : 3 : 4 (recessive epistasis) and 9 : 7 (complementary genes) are modified dihybrid ratios that need gene interaction, which the question does not describe.