Question

1. based on the information in the following table, which man (sammy or george) could possibly be the father of the baby? justify your answer with a punnett square.

name blood type
mother type \ab\
baby type \b\
sammy type \o\
george type \ab\

1. in some chickens, the gene for feather color is controlled by codominance. the allele for black is b and the allele for white is w. the heterozygous phenotype is known as erminette (black and white spotted).

a) what is the genotype for black chickens? ____
b) what is the genotype for white chickens? ____
c) what is the genotype for erminette chickens? ____
sex-linked traits

1. a woman whose father was color blind marries a man who is color blind. what are the chances that their offspring will be color blind? (note: the woman is not color blind.)

a. whats the probability they would have a child that is color blind? ____%
b. whats the probability they would have a child that is not color blind? ____%

Explanation:

Question 11

Step1: Define blood type genotypes

Mother (AB): $I^A I^B$; Baby (B): $I^B I^B$ or $I^B i$; Sammy (O): $ii$; George (AB): $I^A I^B$

Step2: Punnett square (Mother + Sammy)

	$I^A$	$I^B$
$i$	$I^A i$	$I^B i$

Offspring types: A, B. Baby (B) is possible.

Step3: Punnett square (Mother + George)

	$I^A$	$I^B$
$I^B$	$I^A I^B$	$I^B I^B$

Offspring types: A, AB, B. Baby (B) is possible.

Step4: Verify O donor compatibility

Baby can be $I^B i$, getting $i$ from Sammy. George cannot pass $i$, but baby can be $I^B I^B$ (from mother's $I^B$ and George's $I^B$). Both can father the baby.

Codominance means both alleles are expressed in heterozygotes. Homozygous genotypes show the single corresponding trait, while heterozygotes show the mixed erminette phenotype.

Step1: Define genotypes

Color blindness is X-linked recessive ($X^b$). Woman's father is $X^b Y$, so woman is $X^B X^b$ (not color blind). Man is $X^b Y$.

Step2: Punnett square setup

	$X^B$	$X^b$
$Y$	$X^B Y$	$X^b Y$

Step3: Calculate probabilities

Color blind offspring: $X^b X^b$, $X^b Y$ (2 out of 4). Non-color blind: $X^B X^b$, $X^B Y$ (2 out of 4).

Answer:

Both Sammy and George could possibly be the father. Sammy can pass an $i$ allele to produce a baby with genotype $I^B i$ (type B), and George can pass an $I^B$ allele to produce a baby with genotype $I^B I^B$ (type B).

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Question 12