Chapter 4.2, Problem 37AQP

Summary Introduction

To determine:

The probability of obtaining genotype $\text{aa}\text{Bb}\text{Cc}{\text{X}}^{\text{+}}{\text{X}}^{\text{+}}$ in the progeny, if genotype ${\text{Aa Bb Cc X}}^{\text{+}}{\text{X}}^{\text{r}}{\text{ × Aa BB cc X}}^{\text{+}}\text{Y}$ crossed with each other.

Introduction:

Autosomal allele, it may be recessive or dominant. The alleles and their associated genes are autosomal dominant or recessive. X-linked allele mean that the gene is causing genetic disorder is located on the X chromosome in females and Y in males.

Explanation of Solution

Female have two X chromosome and male has one. So the genotype of female is ${\text{Aa Bb Cc X}}^{\text{+}}{\text{X}}^{\text{r}}$ and the male genotype will be ${\text{Aa BB cc X}}^{\text{+}}\text{Y}$. In this a,b, and c are representing autosomal gene.${\text{X}}^{\text{+}}$ and ${\text{X}}^{\text{r}}$ are showing dominant and recessive allele for an X-linked gene.

The probability of obtaining $\text{aa}\text{Bb}\text{Cc}{\text{X}}^{\text{+}}{\text{X}}^{\text{+}}$ will determined by crossing male and female genotype ${\text{Aa Bb Cc X}}^{\text{+}}{\text{X}}^{\text{r}}{\text{ × Aa BB cc X}}^{\text{+}}\text{Y}$.

From Aa genotype of male and female gametes, one half containing A and other half a.

Thus the probability of obtaining genotype aa in the progeny will be:

$\text{1/2a×1/2a=1/4aa}$

In the case of BB genotype: male genotype is BB so, single type of gametes will be produced B.  In females, Bb two type of gametes will produced, one half containing B and other half b. So, probability of obtaining genotype Bb, in the progeny will be:

$\text{1B×1/2=1/2Bb}$

In the case of male cc, genotype it contain single gamete which will produce c only. and in the female Cc two type of gametes will produced half containing C and other half containing c. So, probability of obtaining genotype Cc in the progeny will be:

$\text{1c×1/2C=1/2Cc}$

For sex chromosome: male contain ${\text{X}}^{\text{+}}\text{Y}$ genotype, two type of gametes, half of gamete is containing ${\text{X}}^{\text{+}}$ and other half $\text{Y}$. In female ${\text{X}}^{\text{+}}{\text{X}}^{\text{r}}$ genotype, half of gamete is containing ${\text{X}}^{\text{+}}$ and half containing ${\text{X}}^{\text{r}}$. Thus the probability of obtaining ${\text{X}}^{\text{+}}{\text{X}}^{\text{r}}$ will be:

${\text{1/2X}}^{\text{+}}{\text{×1/2X}}^{\text{+}}{\text{=1/4X}}^{\text{+}}{\text{X}}^{\text{+}}$

Hence, the probability of the genotype $\text{aa}\text{Bb}\text{Cc}{\text{X}}^{\text{+}}{\text{X}}^{\text{+}}$ is

${\text{1/4aa×1/2Bb×1/2Cc×1/4X}}^{\text{+}}{\text{X}}^{\text{+}}\text{=1/64}{\text{aaBbCcX}}^{\text{+}}{\text{X}}^{\text{+}}$

Table 1: Punnet square.

	ABcX${}^{+}$	aBcX${}^{+}$	ABcY	aBcY
ABCX${}^{+}$	AABBCc ${\text{X}}^{\text{+}}{\text{X}}^{\text{+}}$	AaBBCc ${\text{X}}^{\text{+}}{\text{X}}^{\text{+}}$	AABBCc ${\text{X}}^{\text{+}}\text{Y}$	AaBBC ${\text{X}}^{\text{+}}\text{Y}$
ABcX${}^{+}$	AABBcc ${\text{X}}^{\text{+}}{\text{X}}^{\text{+}}$	AaBBcc ${\text{X}}^{\text{+}}{\text{X}}^{\text{+}}$	AABBcc ${\text{X}}^{\text{+}}\text{Y}$	AaBBcc ${\text{X}}^{\text{+}}\text{Y}$
AbCX${}^{+}$	AABbCc ${\text{X}}^{\text{+}}{\text{X}}^{\text{+}}$	AabbCc ${\text{X}}^{\text{+}}{\text{X}}^{\text{+}}$	AABbCc ${\text{X}}^{\text{+}}\text{Y}$	AaBbC ${\text{X}}^{\text{+}}\text{Y}$
AbcX${}^{+}$	AABbcc ${\text{X}}^{\text{+}}{\text{X}}^{\text{+}}$	AaBbcc ${\text{X}}^{\text{+}}{\text{X}}^{\text{+}}$	AABcc ${\text{X}}^{\text{+}}\text{Y}$	AaBbcc ${\text{X}}^{\text{+}}\text{Y}$
aBCX${}^{+}$	AaBBCc ${\text{X}}^{\text{+}}{\text{X}}^{\text{+}}$	aaBBCc ${\text{X}}^{\text{+}}{\text{X}}^{\text{+}}$	AaBBCc ${\text{X}}^{\text{+}}\text{Y}$	aaBBCc ${\text{X}}^{\text{+}}\text{Y}$
aBcX${}^{+}$	AaBBcc ${\text{X}}^{\text{+}}{\text{X}}^{\text{+}}$	aaBBcc ${\text{X}}^{\text{+}}{\text{X}}^{\text{+}}$	AaBBcc ${\text{X}}^{\text{+}}\text{Y}$	aaBBcc ${\text{X}}^{\text{+}}\text{Y}$
abCX${}^{+}$	AaBbCc ${\text{X}}^{\text{+}}{\text{X}}^{\text{+}}$	aaBbCc ${\text{X}}^{\text{+}}{\text{X}}^{\text{+}}$	AaBbCc ${\text{X}}^{\text{+}}\text{Y}$	aaBbCc ${\text{X}}^{\text{+}}\text{Y}$
abcX${}^{+}$	AaBbcc ${\text{X}}^{\text{+}}{\text{X}}^{\text{+}}$	aaBbcc ${\text{X}}^{\text{+}}{\text{X}}^{\text{+}}$	AaBbcc ${\text{X}}^{\text{+}}\text{Y}$	aaBbcc ${\text{X}}^{\text{+}}\text{Y}$
ABCX${}^{\text{r}}$	AABBCc ${\text{X}}^{\text{+}}{\text{X}}^{\text{r}}$	AaBBCc ${\text{X}}^{\text{+}}{\text{X}}^{\text{r}}$	AABBCc ${\text{X}}^{\text{r}}\text{Y}$	AaBBCc ${\text{X}}^{\text{r}}\text{Y}$
ABcX${}^{\text{r}}$	AABBcc ${\text{X}}^{\text{+}}{\text{X}}^{\text{r}}$	AaBBcc ${\text{X}}^{\text{+}}{\text{X}}^{\text{r}}$	AABBcc ${\text{X}}^{\text{r}}\text{Y}$	AaBBcc ${\text{X}}^{\text{r}}\text{Y}$
AbCX${}^{\text{r}}$	AABbCc ${\text{X}}^{\text{+}}{\text{X}}^{\text{r}}$	AabbCc ${\text{X}}^{\text{+}}{\text{X}}^{\text{r}}$	AABbCc ${\text{X}}^{\text{r}}\text{Y}$	AaBbCc ${\text{X}}^{\text{r}}\text{Y}$
AbcX${}^{\text{r}}$	AABbcc ${\text{X}}^{\text{+}}{\text{X}}^{\text{r}}$	AaBbcc ${\text{X}}^{\text{+}}{\text{X}}^{\text{r}}$	AABbcc ${\text{X}}^{\text{r}}\text{Y}$	AaBbcc ${\text{X}}^{\text{r}}\text{Y}$
aBCX${}^{\text{r}}$	AaBBCc ${\text{X}}^{\text{+}}{\text{X}}^{\text{r}}$	aaBBCc ${\text{X}}^{\text{+}}{\text{X}}^{\text{r}}$	AaBBCc ${\text{X}}^{\text{r}}\text{Y}$	aaBBCc ${\text{X}}^{\text{r}}\text{Y}$
aBcX${}^r$	AaBBcc ${\text{X}}^{\text{+}}{\text{X}}^{\text{r}}$	aaBBcc ${\text{X}}^{\text{+}}{\text{X}}^{\text{r}}$	AaBBcc ${\text{X}}^{\text{r}}\text{Y}$	aaBBcc ${\text{X}}^{\text{r}}\text{Y}$
abCX${}^{\text{r}}$	AaBbCc ${\text{X}}^{\text{+}}{\text{X}}^{\text{r}}$	aaBbCc ${\text{X}}^{\text{+}}{\text{X}}^{\text{r}}$	AaBbCc ${\text{X}}^{\text{r}}\text{Y}$	aaBbCc ${\text{X}}^{\text{r}}\text{Y}$
abcX${}^{\text{r}}$	AaBbcc ${\text{X}}^{\text{+}}{\text{X}}^{\text{r}}$	aaBbcc ${\text{X}}^{\text{+}}{\text{X}}^{\text{r}}$	AaBbcc ${\text{X}}^{\text{r}}\text{Y}$	aaBbcc ${\text{X}}^{\text{r}}\text{Y}$

Conclusion

X-linked allele mean that the gene is causing genetic disorder is located on the X chromosome in females and Y in males. The genotype of female is ${\text{Aa Bb Cc X}}^{\text{+}}{\text{X}}^{\text{r}}$ and the male genotype will be ${\text{Aa BB cc X}}^{\text{+}}\text{Y}$. The probability of obtaining $\text{aa}\text{Bb}\text{Cc}{\text{X}}^{\text{+}}{\text{X}}^{\text{+}}$ is determined by crossing male and female genotype ${\text{Aa Bb Cc X}}^{\text{+}}{\text{X}}^{\text{r}}{\text{ × Aa BB cc X}}^{\text{+}}\text{Y}$. The probability of the genotype $\text{aa}\text{Bb}\text{Cc}{\text{X}}^{\text{+}}{\text{X}}^{\text{+}}$ is $\text{1/64}{\text{aaBbCcX}}^{\text{+}}{\text{X}}^{\text{+}}$