Question

i need help with steps please

Question 1 (1 point) A woman who is homozygous for type A blood, mates with a heterozygote who has type B blood. What is the probability that their child will have type A blood? O 0% O25% 50% 75% 100% Question 2 (1 point) Red-green colourblindness is a sex-linked trait. If a male with normal vision mates with a female carrier, which of the following is correct? O100% of their daughters will be carriers O100% of their sons will be red-green colourblind 50% of their daughters will be red-green colourblind 50% of their daughters will be carriers 50% of their sons will be carriers

Answer 1

1) three alleles at the ABO locus determines the blood group

IA, IB, i

genotype of the woman who is homozygous for type A blood is IAIA and genotype of the man who is heterozygote for the type B blood.

IAIA * IBi

pe

	IB	i
IA	IAIB ( AB blood type)	IAi (A blood type)
IA	IAIB ( AB blood type)	IAi (A blood type)

percentage of progeny with A blood type= ( number of progeny with A blood type/total number of progenies)$\times$100

= (2/4)$\times$100

= 50%

so the answer is 50%.

2) let the alleles be XC- normal vision Xc- colorblindness

then genotype of the man with normal vision is XCY ( males are XY they have only one X chromosome)

the genotype of the heterozygous female is XCXc

XCXc * XCY

	XC	Xc
XC	XCXC (normal female)	XCXc ( carrier female)
Y	XCY ( normal male)	XcY (colorblind male)

percentage of carrier daughters= ( number of carrier daughters/total number of daughters )$\times$100

= ( 1/2)$\times$100

= 50%

here half of the sons are affected and half are normal, males cannot be carriers for the disease, they have only one X chromosome, allele in that X chromosome will be expressed, heterozygous females are carriers.

so the answer is 50% of the daughters will be carriers