Chapter 11, Problem 8A

Summary Introduction

To predict:

The genotypes of the children of a father with Huntington’s disease and an unaffected mother

Introduction:

Punnett developed a method to predict the possible offspring of a cross between two known genotypes. It is easier to keep a track of the possible genotypes involved in a cross. Huntington’s disease is a dominant genetic disease that affects the nervous system.

Answer to Problem 8A

Case I- Father is heterozygous for dominant gene and mother is unaffected.

Based on the following Punnett square,there are 50% offspring who have the disorder and 50% who do not have the disorder. The genotypes of the children will be 50% heterozygous and 50% homozygous for recessive gene.

	Y	y
y	Yy	yy
y	Yy	yy

Case II- Father is homozygous for dominant gene and mother is unaffected.

Based on the following Punnett square all offspring have the disorder. The genotypes of all the children will be heterozygous for the gene.

	Y	Y
y	Yy	Yy
y	Yy	Yy

Explanation of Solution

Huntington’s disease is a dominant genetic disorder so it will be expressed in both homozygous (for dominant gene) and heterozygous condition. An unaffected person will be homozygous recessive for the gene.

There can be two cases; one in which the father is heterozygous for the dominant gene and another in which father is homozygous for the gene. Since mother is unaffected, she will be homozygous recessive for the gene.

Case I A heterozygous father can be represented as Yy .

An unaffected mother can be represented as yy.

The children will have only two possible genotypes; Yy and yy .

This can be checked by drawing a Punnett square as shown below.

	Y	y
y	Yy	yy
y	Yy	yy

So, the number of children with homozygous genotype is 2 and heterozygous genotype is also 2. Hence,50% children will be affected by the disease and 50% will not be affected by disease.

Case II A homozygous father can be represented as YY .

An unaffected mother can be represented as yy.

The children will have only one possible genotypes; Yy .

This can be checked by drawing a Punnett square as shown below.

	Y	Y
y	Yy	Yy
y	Yy	Yy

So, all children will have heterozygous genotype. Hence,all children will be affected by the disease.