Question

1. Practice with pedigree charts. A man and his maternal grandfather have a particular condition. This man married a woman and had four children in the following order: boy, boy, girl, boy. All the children of this marriage except the oldest boy had the condition. The woman of this marriage had a paternal grandfather that had the condition. Only the above individuals have the condition. a. Draw a pedigree showing the above information. Use appropriate pedigree symbols. b. Is the condition dominant or recessive? c. Write genotypes for all individuals you can in the pedigree. d. What is the probability that the oldest son of the last generation will have a child with the condition if he marries a woman: (1) with the condition (2) who does not have the condition, but is a carrier of the condition (3) without the condition and who is not a carrier of the condition Note: You will need to construct 3 different Punnett squares. You may have to take a picture of this page and upload separately.

Answer 1

a.

N 0 I 고 . 3 TU 2. TV 2 3. J

b.

The condition is recessive. This is because the Man is affected, however, neither of his parents are. This is only possible when the condition is Autosomal Recessive.

c.

If the dominant allele is represented by A, and the recessive allele is represented by a:

Individual	Genotype
I-1	aa
I-2	A_ (This individual can be of the genotype AA or Aa)
I-3	aa
I-4	A_
II-1	A_
II-2	Aa
II-3	Aa
II-4	A_
III-1	aa
III-2	Aa
IV-1	Aa
IV-2	aa
IV-3	aa
IV-4	aa

d)

The first son of this couple is a carrier for the condition. Thus the probability of him passing on the recessive allele is 1/2.

1) With the condition:

If the woman this child marries has the condition, she is certain to pass on the 'a' allele. Thus,

The probability of having an affected child = 1/2 * 1 = 1/2

		Gametes of woman with the condition
		a	a
Gametes of IV-1	A	Aa	Aa
a	aa	aa

2) Who does not have the condition but is a carrier of the condition:

If the woman this child marries is a carrier, she will pass on the 'a' allele with a probability of 1/2. Thus.

The probability of having an affected child = 1/2 * 1/2 = 1/4

		Gametes of woman who does not have the condition but is a carrier
		A	a
Gametes of IV-1	A	AA	Aa
a	Aa	aa

3) Who does not have the condition and is not a carrier of the condition:

Since the woman this child marries is not a carrier, the probability that she passes on the 'a' allele is 0. Thus,

The probability of having an affected child = 1/2 * 0 = 0

		Gametes of woman who does not have the condition and is not carrier
		A	A
Gametes of IV-1	A	AA	AA
a	Aa	Aa