Back to homeBiology II · High School Year 3 · Physics-Math · Change subject
Start for free
Prep 3 Tech · Biology II
Genetics and evolution · Genetics and inheritance

Dominance and Recessiveness

In a family in Oaxaca, the father and mother both have dark hair, and their daughter is born with light hair. Where did that trait come from if neither parent shows it? More than 150 years ago, Gregor Mendel patiently counted thousands of pea plants and found the answer in the numbers.

Gene, allele, genotype, and phenotype

On an ear of corn from a milpa, white kernels can appear among purple kernels, while on other ears all the kernels look the same. Where is that information stored, and how can a trait be passed on when the mother plant does not show it?

Pair of homologous chromosomes, one maternal and one paternal; at the same locus, the maternal one carries allele A and the paternal one carries allele a, so the genotype is Aa. gene locus allele A allele a maternal chromosome paternal chromosome homologous pair · genotype Aa
Practice★☆☆Not started
Define the terms gene, allele, genotype and phenotype, and provide a simple example relating them (for instance, flower color).

Define the terms gene, allele, genotype and phenotype, and provide a simple example relating them (for instance, flower color).

Solve

Dominance, homozygosity, and heterozygosity

Three genotypes, AA, Aa, and aa: the first two give a flower with the dominant phenotype and the third gives a flower with the recessive phenotype. AA homozygous dominant phenotype Aa heterozygous dominant phenotype aa homozygous recessive phenotype AA and Aa: same phenotype, different genotype
Practice★☆☆Not started
Explain what it means for an allele to be dominant and for an allele to be recessive. Can a recessive trait be expressed in a heterozygous individual? Reason your answer.

Explain what it means for an allele to be dominant and for an allele to be recessive. Can a recessive trait be expressed in a heterozygous individual? Reason your answer.

Solve
Practice★☆☆Not started
The diagram shows a Punnett square for a cross between two heterozygous individuals (Aa × Aa). Complete the missing cell and, assuming allele A (dark color) is dominant…

The diagram shows a Punnett square for a cross between two heterozygous individuals (Aa × Aa). Complete the missing cell and, assuming allele A (dark color) is dominant over a (light color), state the genotypic and phenotypic ratios of the offspring.

A a A a AA Aa Aa ?
Solve

Mendel's laws: uniformity and segregation

Mendel crossed pure yellow pea plants with pure green ones, and the whole F₁ came out yellow. When he crossed those plants with each other, he counted 6,022 yellow and 2,001 green (about $75\,\%$ and $25\,\%$). How did green reappear if it had disappeared completely?

Practice★★☆Not started
In pea plants, the allele for yellow color (A) is dominant over green color (a). Cross two heterozygous plants and state the expected genotypic and phenotypic ratios in…

In pea plants, the allele for yellow color (A) is dominant over green color (a). Cross two heterozygous plants and state the expected genotypic and phenotypic ratios in the offspring.

Solve
Practice★★☆Not started
In dogs, black coat (N) is dominant over brown coat (n). A heterozygous male is crossed with a brown-coated female. Calculate the probability that a puppy will have a…

In dogs, black coat (N) is dominant over brown coat (n). A heterozygous male is crossed with a brown-coated female. Calculate the probability that a puppy will have a brown coat.

Solve

Probability and the test cross

A couple who are carriers of a recessive disease have a first child who is affected and fear a second pregnancy, thinking the risk is now higher. Does one pregnancy influence the next?

4-by-4 grid with the 16 equally likely genotype combinations for two children of an Aa by Aa couple; only one, aa in both, corresponds to 1/4 times 1/4 equals 1/16. 2nd child 1st child AA Aa aA aa AA Aa aA aa aa, aa P(aa) = 1/4 P(aa) = 1/4 P(both aa) = 1/4 × 1/4 = 1/16 16 equally likely combinations
Practice★★☆Not started
A couple, both with the dominant phenotype for a trait, have a child with the recessive phenotype. Deduce the genotypes of the parents and explain how it is possible for…

A couple, both with the dominant phenotype for a trait, have a child with the recessive phenotype. Deduce the genotypes of the parents and explain how it is possible for the child to display a trait that neither parent shows.

Solve
Practice★★★Not started
In rabbits, black fur ($N$) is dominant over white fur ($n$). A breeder crosses a black male of unknown genotype with a white female to perform a test cross. A litter of…

In rabbits, black fur ($N$) is dominant over white fur ($n$). A breeder crosses a black male of unknown genotype with a white female to perform a test cross. A litter of 6 kits is born, all with black fur. The breeder concludes: "It is 100% proven that the male is homozygous ($NN$), because if he were heterozygous, white rabbits would have been born".

  1. a) Refute the breeder's conclusion by calculating the probability that a litter of 6 kits is entirely black even if the father is heterozygous ($Nn$).
  2. b) Explain what minimum outcome in the litter would have been sufficient to determine the male's genotype beyond any doubt, and justify why.
Solve
Practice★★★Not started
Albinism is an autosomal recessive condition ($a$) compared to normal pigmentation ($A$). A couple in which both have normal pigmentation have a firstborn child with…

Albinism is an autosomal recessive condition ($a$) compared to normal pigmentation ($A$). A couple in which both have normal pigmentation have a firstborn child with albinism. Facing a new pregnancy, the father-to-be comments: "Since the probability of albinism is 1 in 4 and the first child was already born with albinism, the next three will definitely be born with normal pigmentation."

  1. a) Refute the father's statement by calculating the actual probability that the second child is born with albinism and the probability that the next three are all born with normal pigmentation.
  2. b) If a daughter with normal pigmentation is eventually born, calculate with justification the probability that she is a carrier of the albinism allele.
Solve

Dihybrid inheritance and the third law

Dihybrid cross AaLl by AaLl: the 3:1 ratios of each gene combine at random and give the 9:3:3:1 ratio. AaLl × AaLl (two independent genes) color gene 3 yellow : 1 green texture gene 3 smooth : 1 wrinkled combine at random 9 yellow-smooth : 3 yellow-wrinkled : 3 green-smooth : 1 green-wrinkled 9 : 3 : 3 : 1 ratio
Practice★★★Not started
In peas, yellow color ($A$) is dominant over green ($a$), and smooth texture ($L$) is dominant over wrinkled ($l$). After self-fertilizing dihybrid plants ($AaLl$), a…

In peas, yellow color ($A$) is dominant over green ($a$), and smooth texture ($L$) is dominant over wrinkled ($l$). After self-fertilizing dihybrid plants ($AaLl$), a farmer selects only plants with green, smooth seeds for commercial cultivation, but needs to ensure they are true-breeding ($aaLL$) so that wrinkled seeds do not appear in subsequent harvests.

  1. a) Design a cross that allows this farmer to find out whether a green, smooth plant is homozygous or heterozygous for texture, indicating the genotype and phenotype of the plant used as a mate.
  2. b) Predict with reasoning the phenotypic percentages that would be obtained in the offspring according to each of the possible genotypes, and conclude which outcome would confirm that the plant is true-breeding.
Solve

Back to the beginning: two dark-haired parents can have a light-haired daughter if both are heterozygous ($Aa$): each contributes $a$ with probability $1/2$, and the daughter is $aa$ with probability $1/4$ ($25\,\%$).

Summary

See all the exercises in this section