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.

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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?

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A bald man (C-) whose father was not bald (cc) ... the man has to be Cc, because his father was not bald.

... he had children with a woman who is not bald (C-), but whose mother was (cc) ... the woman will be Cc, because her mother was bald.

Normal grandfather Bald grandmother

XY cc XX CC

Bald Man x Normal Woman

Gametes XY Cc XX Cc

	XC	X c	YC	And c
XC	XX CC	XX Cc	XY CC	XY Cc
X c	XX Cc	XX cc	XY Cc	XY cc

1/4 of the daughters will be bald and 3/4 of the male children will be bald.

We could also do it without using the sex chromosomes:

Crossing: Cc x Cc

Gametes: C c C c

F1: CC Cc Cc cc

Men: 75% bald 25% hairy

Women: 25% bald 75% with hair

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.

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First we will have to remember how the baldness gene works in humans, being dominant in men and recessive in women, in such a way that the possible genotypes are manifested with these phenotypes, according to sex:

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

We will try to reread the story by deducing the genotypes.

Antonio, Paquirri's father was bald (C_) had two children who were not bald, neither Paquirri (cc) nor José (cc), so Antonio's genotype must be Cc.

Paquirri (cc) had two children with Carmina Ordóñez, who was not bald (c_), Fran and Cayetano, both with hair, so they have to be (cc).

On the other hand, Carmina (c_) had another son with Julián Contreras, who has hair (cc), Julián Jr., also with hair (cc). Carmina had three children, and all three had the allele c, so although it is more likely that it was cc, we cannot rule out that it was Cc.

Paquirri (cc) had a relationship with Isabel Pantoja , who has hair (c_), Paquirrín, who is bald (Cc). As Paquirrín is bald (C_) and her father was not (cc), Isabel has to be (Cc) in order not to be bald and have a bald son (Cc).

Juan, Isabel's father was also bald (C_), but neither she (Cc) nor her three brothers, Agustín (cc), Juan Antonio (cc) nor Bernardo (cc), are bald. Therefore, Juan has to be Cc.

Summarizing:

Antonio: Cc, bald.
Paquirri: cc, with hair.
José: cc, with hair.
Carmina: cc or Cc, with hair.
Fran: cc, with hair.
Cayetano: cc, with hair.
Julian: cc, with hair.
Julián Jr.: cc, with hair.
Isabel: Cc, with hair.
Paquirrín: Cc, with hair.
Juan: cc, with hair.
Augustine: cc.
Juan Antonio: cc.
Bernardo: cc.

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?

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Parents White dog _____________ ____________ Black dog

BB ? ? NN

The father of the dog, the one with white hair, has to be BB because white is recessive in males.

The mother of the bitch, with black hair, has to be NN because B and N have intermediate inheritance in females.

The black dog in the statement, the son of the white dog, must be NB, heterozygous.

The gray bitch, daughter of the black bitch, will have to be NB, heterozygous, due to the intermediate dominance.

Parents White dog _____________ ____________ Black dog

BB ? ? NN

P NB x NB

Gametes N B N B

F1 NN NB NB BB

NB genotypes: 50%. Black dog, gray bitch.

NN: 25%. Black dog, black bitch.

BB: 25%, White dog, white bitch.

75% of the males of the offspring will be black. The remaining 25%, white.

50% of the female offspring will be gray, 25% white, and the other 25% black.

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 X^d: color blindness, determine:

a) The genotypes of men and women.

b) Genotypes and phenotypes of the first filial generation.

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a) Genotypes of men and women.

This is a somewhat more complex problem than the ones we have done so far. That is why we will have to raise it with a little more care.

Two different characters appear in the problem:

Colour blindness:
- Men:
  - Normal vision: XY.
  - Colorblind: X^dY
- Women:
  - Normal vision: XX, X^dX.
  - Color blind: X^dX^d.
Baldness:
- Men:
  - Calvos: CC, CN.
  - Furry: NN.
- Women:
  - Calvas: CC.
  - Hairy: CN, NN.

With these data present, we can now begin to pose the problem.

Heterozygous bald and color blind man: CN X^dY .
Woman without baldness and with normal color vision, whose father was neither color blind nor bald and whose mother was bald and with normal vision (not a carrier of the gene for color blindness). Since the woman is not bald, she has to be _N, but since her mother was bald (CC), she will be CN. She is not color blind, her father was not color blind (XY) and neither was her mother (XX^d or XX), so in the absence of more information, it will be XX or X^d). As they then ask us about their descendants, we will think that it is CNXX .

b) Genotypes and phenotypes of the first filial generation.

CNX^dY x CNXX

Gametes CX^d CY NX^d NY x CX NX

	CX^d	CY	NX^d	NY
CX	CCXX^d	CCXY	CNXX^d	CNXY
NX	CNXX^d	CNXY	NNXN^d	NNXY

The genotypes and phenotypes of the first filial generation are:

Bald spots and carriers of color blindness: CCX^dX.
Hairy and color blind carriers: CNX^dX
Hairy and color blind carriers: CNX^dX
Hairy and color blind carriers: NNX^dX
Bald with normal vision: CNXY
Bald with normal vision: CNXY
Bald with normal vision: CCXY
Bald with normal vision: CNXY
Hairy with normal vision: NNXY

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.

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A bald man (XYC_) whose father (XYcc) and mother (XXC_) are not bald, and a woman who is not bald (XX C_) whose father (XYC_) and mother (XX cc) are, want to have children:

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

Grandparents XYcc x XXC_ XYC_ x XXcc

Pair XYCc XXCc

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

Pair XYCc x XXCc

Gametes XC Xc YC Yc XC Xc

Possible genotypes:

1/8 XXCC hairy women

2/8 XXCc hairy women

1/8 XXcc bald women

1/8 XYCC bald men

2/8 XYCc bald men

1/8 XYcc hairy men

1/4, 25% of daughters will be bald.

3/4, 75% of the children will be bald.

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