Question

Chris (Blood Type A; Widows Peak (an autosomal dominant trait), normal skin pigment) and his wife Pat (Blood Type B; No Widows Peak; Albino (an autosomal recessive trait) have four children. Chris has a sister with cystic fibrosis (an autosomal recessive trait), but Chris appears normal. Chris is colorblind (an X-linked recessive trait), but Pat is not.

Which of the following might NOT be the child of Chris and Pat?

Blood Type A; Widows Peak; Normal Skin Pigment

Blood Type B; No Widows Peak; Albino

Blood Type AB; Widows Peak; Normal Skin Pigment

Blood Type O; No Widows Peak; Albino

Any of them could be their child.

If Chris is heterozygous for skin pigment; what expected percent of potential offspring with Pat would be albino?

		0?
		25%
		50%
		75%
		100% If Chris and Pat’s first child is a daughter that is colorblind. They have another daughter, what are the chances that she is color-blind? (Draw a Punnett square if needed) 0 1/16 1/4 1/2 1

Answer 1

1) ABO locus determines blood, IA, IB, and I are three alleles that determine the blood group.

IAIA,IAI- A-type blood, IAIB- AB type blood, IBIB, IBI- B type blood, II- O type blood.

let the alleles be W- Widow`s peak, w- straight hairline.

A- normal skin pigment, a- albinism

Chris (Blood Type A; Widows Peak (an autosomal dominant trait), normal skin pigment) so his genotype can be IAIWwAa and Pat (Blood Type B; No Widows Peak; Albino (an autosomal recessive trait) her genotype can be IBIwwaa.

IAIWwAa * IBIwwaa

IAI * IBI

	IA	I
IB	IAIB ( AB type blood)	IBI ( B type blood)
I	IAI ( A type blood)	II ( O type blood)

Ww * ww

	W	w
w	Ww ( widow`s peak)	ww( straight hair line)
w	Ww ( widow`s peak)	ww( straight hair line)

Aa * aa

	A	a
a	Aa ( normal)	aa ( albino)
a	Aa ( normal)	aa ( albino)

so they can have progenies with

Blood Type A; Widows Peak; Normal Skin Pigment

Blood Type B; No Widows Peak; Albino

Blood Type AB; Widows Peak; Normal Skin Pigment

Blood Type O; No Widows Peak; Albino

so the answer is Any of them could be their child.

2) Chris is heterozygous for pigmentation

Aa * aa

	A	a
a	Aa ( normal)	aa ( albino)
a	Aa ( normal)	aa ( albino)

Percentage of albino progeny=( number of aa/total number) 100=(2/4)100=50%

b) 50%

3) Chris is colorblind (an X-linked recessive trait), but Pat is not.

Colourblindness is sex-linked recessive.

let the alleles be XC- normal and Xc-colourblind

Chris and Pat’s first child is a daughter that is colorblind, so Pat is Heterozygous.

The genotype of Pat is XCXc and the genotype of Chris is XcY.

XCXc * XcY

	XC	Xc
Xc	XCXc	XcXc
Y	XCY	XcY

having a child is an independent event it does not depend on the phenotype of elder children.

Probability that a girl child is colourblind= number of colourblind daughters/total number of daughters=1/2