Question

A Punnett is a means to determine the genetic inheritance of offspring if the genotypes of both parents are known. Using Punnett squares answer the questions about the following scenarios. In order to properly answer some of the questions more than 1 Punnett square might be needed. With every Punnett square provide a key for your alleles.

1. Huntington’s disorder is caused by a dominant allele. Heterozygous individuals with one copy of this allele suffer from progressive degeneration of the nervous system and premature death. The phenotype or symptoms do not appear until the affected individual reaches ~40 years of age. This is after child-bearing age, so people have children before they know they are carrying the affected gene allele.

Jermaine is very worried about passing the dominant Huntington’s gene allele to his children. If Jermaine’s father has the Huntington’s Disorder, his mother does not, and his wife has no history of Huntington’s, does Jermaine need to be tested for Huntington’s before he has children? Note: homozygous dominant is lethal and a fetus will not survive. Draw a punnett square representing the Jermaine’s father and mother, to determine his chances of carrying the Huntington’s disease allele. (H = dominant, disease allele; h = recessive, non-disease allele)

What is the probability that Jermaine has the Huntington’s disease allele? Should he be tested?

If Jermaine is positive for the Huntington’s allele, what are the chances he will pass it on to his children? Do a second punnett square to illustrate this cross.

Answer 1

Jermaine's father is a heterozygote and his mother is homozygous recessive ( his father is affected and his mother is not affected)

let the alleles be H-Huntington disease h-normal

Jermaine`s father's genotype is Hh and Jermaine`s mother is hh.

Hh * hh

	H	h
h	Hh ( Huntington`s disease)	hh (normal)
h	Hh ( Huntington`s disease)	hh (normal)

the probability that Jermaine has Huntington`s disease allele= number of Hh individuals/total number of progenies=2/4=1/2

the probability that Jermaine has Huntington`s disease allele is 1/2 so it is better he goes for testing.

If Jermaine is posiitve for the H allele, his genotype is Hh.

His wife is hh because she does not have a history of Huntington`s disease.

Hh * hh

	H	h
h	Hh ( Huntington`s disease)	hh (normal)
h	Hh ( Huntington`s disease)	hh (normal)

chances Jermaine will pass it on to his children= (number of Hh progenies/total number of progenies)100

= (2/4)100

= 50%