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3.5.5. Sex-influenced inheritance (problems solved)

Heredity influenced by sex (problems solved). (Page 22)

As we have seen when we talk about inheritance influenced by sex, these are characters determined by genes that are not on the sex chromosomes, but whose manifestation depends on the sex of the individual. It is not linked to sex, but it is influenced by sex .

If a character has a different type of inheritance depending on whether it is men or women, there are alleles that behave as dominant in men and recessive in women, or vice versa.

To resolve this type of inheritance, sex must be considered as one character, and act as if it were the inheritance of two characters (as in Mendel's third law).

The most typical case is that of baldness in humans. The allele that determines baldness is dominant in men but recessive in women. Thus, the heterozygous genotype in men will lead to a bald person, while in women it will lead to a non-bald person.

  • CC genotype: bald men and bald women.
  • Genotype Cc: bald men, women with hair.
  • Genotype cc: men and women with hair.

Genetics problem 161

A bald man whose father was not bald had children with a woman who is not bald, but whose mother was. How likely are your children to be bald?

Genetics problem 162

Antonio, Paquirri's father was bald, but neither Paquirri nor his brother José were bald. Paquirri had two children with Carmina Ordóñez (he had hair), Fran and Cayetano, both handsome and with hair. Later, Carmina had another son with Julián Contreras (with hair), Julián Jr., also with hair, and Paquirri had another son, Paquirrín, with Isabel Pantoja, nice but bald and who likes to be called Kiko Rivera. Juan, Isabel's father was also bald, but neither she nor her three brothers, Agustín, Juan Antonio, nor Bernardo, are bald.

Deduce the genotype of all family members.

Paquirri                         Isabel Pantoja

Hacerlo en educaplay.

Genetics problem 163

A black-haired dog, whose father was white, is crossed with a gray-haired female dog, whose  mother was black. We know that black fur dominates over white in males, and that in females black and white have intermediate inheritance. Use the letters N and B as symbols to represent the allele that gives rise to black and white respectively.

a) How will the genotypes of the dogs that are crossed?
b) What will be the phenotype and genotype of their offspring?
c) What does dominance, recessivity and intermediate inheritance mean?

Genetics problem 164

A recessive sex-linked gene causes color blindness in men. A gene influenced by sex determines baldness (dominant in males and recessive in females). A heterozygous bald and color-blind man marries a woman without baldness and normal color vision, whose father was neither color blind nor bald and whose mother was bald and normal vision (not a carrier of the color blind gene). Using the nomenclature C: baldness and N: no baldness, and X: normal vision and Xd: color blindness, determine:

a) The genotypes of men and women.

b) Genotypes and phenotypes of the first filial generation.

Genetics problem 165

The D1 allele determines an index finger shorter than the ring finger, it is dominant in men but recessive in women. The D2 allele determines an index finger equal to or longer than the ring finger, it is recessive in men and dominant in women.

a) What would be the type of offspring expected of a man with a long index finger and a woman with a short index finger?
b) What type of inheritance is it?

Genetics problem 166

Premature baldness has a sex-influenced inheritance (not linked to sex chromosomes). The allele that determines baldness (C) is dominant in men, but recessive in women (c). Thus, the heterozygous genotype in men will lead to a bald person, while in women it will lead to a non-bald person. A bald man whose father and mother are not bald, and a non-bald woman whose father and mother are, wish to have children:

a) Indicate the genotypes of the six individuals mentioned in the statement;

b) What is the probability that their descendants are bald? Indicate the probability separately for daughters and sons.