Mendelian Principles Questions and Answers

Q: A female cat with a mutant phenotype was bred with a wild-type male cat. All progeny (4 males and 4 females) show the mutant phenotype. On the other hand, all progeny (4 males and 4 females) from the reciprocal cross between a mutant male and a wild-type female show the wild-type phenotype. Which of the following explain the inheritance pattern of the mutation?

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.

Q: Mendel deduced the genetic principle of inheritance by experimenting on sweet pea plants. One of the experiments involved crossing plants with two contrasting characters, tall (dominant) and dwarf (recessive), which yielded all tall plants in the first generation. When the same genetic cross was independently repeated by a researcher, only short plants were obtained. Which one of the following can possibly explain the altered outcome?

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.

Q: What would be the number of genotypes and phenotypes, respectively, from a cross between genotypes AaBBCcDd and AaBBCcDd ? Assume independent assortment and simple dominant-recessive relationship in each gene pair.

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.

A female cat with a mutant phenotype was bred with a wild-type male cat. All progeny (4 males and 4 females) show the mutant phenotype. On the other hand, all progeny (4 males and 4 females) from the reciprocal cross between a mutant male and a wild-type female show the wild-type phenotype. Which of the following explain the inheritance pattern of the mutation?

A.Recessive
B.Linked inheritance
C.Mitochondrial inheritance
D.Autosomal inheritance

Exam Year(s)

GATE Life Sciences - 2012,

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.

Discuss in Forum

Mendel deduced the genetic principle of inheritance by experimenting on sweet pea plants. One of the experiments involved crossing plants with two contrasting characters, tall (dominant) and dwarf (recessive), which yielded all tall plants in the first generation. When the same genetic cross was independently repeated by a researcher, only short plants were obtained. Which one of the following can possibly explain the altered outcome?

A.Tall plants were heterozygous
B.An enhancer for the tall allele is present in the dwarf plant
C.A suppressor for the tall allele is present in the dwarf plant
D.Dwarf plants are homozygous

Exam Year(s)

GATE Life Sciences - 2022,

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.

Discuss in Forum

What would be the number of genotypes and phenotypes, respectively, from a cross between genotypes AaBBCcDd and AaBBCcDd ? Assume independent assortment and simple dominant-recessive relationship in each gene pair.

A.8 and 4
B.12 and 4
C.27 and 8
D.14 and 8

Exam Year(s)

GATE Life Sciences - 2025,

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.

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