punnett square problem?

Question

two parents with normal color perception have three children, two girls and one boy. The girls have normal color perceptin but the boy is red-green colorblind. Use punnett square to show the resulting genotypes and phenotypes of the parents and offspring.



a patertnity case involves these facts: the woman is blood type A+, and her child is blood type o-, and the alleged father is blood type b+. could he be the father? Explain using a puneet square.

Answer 1

Because red-green colorblindness is a sex-lined recessive trait on the X-chromosome:

1. A boy with colorblindness has genotype Xb Y.
He got Y from Dad and Xb from Mom.

2. Dad does not have colorblindness, so he is XB Y.

3. Mom has a colorblind son but is not colorblind herself. Her genotype is XB Xb.

4. The daughters can either be XB Xb or XB XB.

Make the Punnett square with alleles from Mom on the left XB and Xb. Put the alleles from Dad on the top XB and Y. Fill in the Punnett square in the usual way.

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Yes, he can be the father. Here's the reasoning.
1. The child is Type O negative. This means that the child got the O allele from both parents, as well as the Rh negative allele from both parents.

2. Mom is blood type A+, so her genotype must be AO, +/-.
3. Dad is blood type B+, so his genotype must be BO, +/-.

A Punnett square can be made with the gametes from the mother on the left: A+, A-, O+, O-. Put the gametes from the suspected father on the top of the square: B+, B-, O+, O-.

When you fill in the results, the possible genotype for the Type O negative child is in the third and fourth rows, in the last box of each row.

There is a 2/16 or 1/8 chance of these two parents have a child with Type O negative blood.

Answer 2

OK, I can't really draw out Punnett squares here, but I'll explain as best I can.
1. If I remember correctly, red-green colorblindness is a recessive sex-linked trait. Meaning if both parents have normal color perception, the father has the dominant allele on his X chromosome (XY), and the mother is either homozygous dominant (XX) or heterozygous (Xx). Since the son is colorblind, the mother must be heterozygous. The offspring would then be: Son recessive (xY), daughters either homozygous dominant (XX) or heterozygous (Xx)
2. Yes, the alleged father COULD be the father. O type blood and negative Rh factor blood are both recessive traits. In order for the child to be related to this alleged father, the mother would have to be (AO + -), and the father would be (BO + -). That way, they would both give the O alleles and the negative Rh factor alleles to the offspring, resulting in an O- blood type.

Answer 3

Since colorblindness is X-linked recessive...
Normal = X-C, Colorblind = X-c.
The mother has to be the carrier, as the father can't be. he only carries one copy of that allele, so if that one was abnormal, he'd then be colorblind. The mother is a carrier, and it doesn't show up because she has a dominant normal allele present, too. Therefore, the parental genotypes are mother: X-C/X-c, and father, X-C/Y.

____X-C______Y
X-C || X-C/X-C || X-C/Y
X-c ||X-C/X-c || X-c/Y

X-C/X-c, X-C/X-C are normal girls. The first is a carrier, the second doesn't even carry the colorblindness allele.

X-C/Y is a normal boy.

X-c/Y is a colorblind boy.

Now, for the second question. It is possible for the man to be the father. Just imagine that the mom is AO and the dad is BO, and both are R1r (or heterozygous for the rhesus protein). This makes it possible for the child to still be O-.

____B+__B-___O+___O-
A+ || AB+ || AB+ || AO+ || AO+
A- || AB+ || AB- || AO+ || AO-
O+ || BO+ || BO+|| OO+|| OO+
O- || BO+ BO- || OO + || OO-

Of course, as you can tell, the chances are still pretty slim of that happening (1/16)

Answer 4

were learning about punnet squares in skool now!! lol!!