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Genotype/Phenotype Help--

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Genotype/Phenotype Help--

Postby envii » Wed Apr 06, 2005 10:38 pm

I have a biology exam tomorrow on genetics, and I'm trying to complete a study-guideish set of questions we were given. However, I have no idea what I am doing! :oops:

The question I'm trying to figure out right now is as follows:

Hemophilia is caused by a sex-linked recessive allele (x^h). Polydactylism is due to a dominant autosomal allele (P). A polydactyl man who is not a hemophiliac has a child with a woman who has the normal phenotype for both of these traits. Their first child is a male who is normal for both of these traits. The second child is a polydactyl boy with hemophilia.

a) What are the genotypes of both children?
b) What are the genotypes of the parents?
c) What is the chance that they have a female child who is normal for both traits?


Now, we weren't taught how to combine autosomal and sex-linked genes into one genotype. I don't know if the father would be PX^hY and the mother X^h+Xh or what... Can someone please help? :?
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Postby biostudent84 » Wed Apr 06, 2005 11:23 pm

Working through this problem one disease at a time.

Hemophilia:

Non-hemophilic father: XY
Normal* mother: XX orX(X^h)

We know that at least one of the mother's X chromosomes is normal, but we do not know for sure that they both are normal. Only one normal X chromosome is required to prevent hemophilia from being expressed. The other could or could not be an X^h

Child 1: Normal: XY
Child 2: Hemophiliac: (X^h)Y

The second child's (X^h) had to have come from somewhere, so therefore, we now know that the mother is X(X^h).

Polydactyly:

Polydactyl father: Pp or PP
Normal mother: pp

The mother being normal, and the trait being dominant, we know she has no gene for polydactyly. The father, having polydactyly, we know that at least one of his alleles has the gene...it could be one allele, it could be both.

Child 1: Normal: pp
Child 2: Polydactyly: Pp

Child 2 must be Pp because one comes from the mother (which will always be P), and the other from the father. And as the child must get a P from somewhere, it can only come from his polydactylic father.

Now we can look at the father again. Since each parent gives one allele to each child, we know from Child 1 that the father has at least one p. Combining this with information above, we now know that the father must be Pp.

Father: XYPp
Mother: X(X^h)pp
Child 1: XYpp
Child 2: (X^h)YPp

If you combine father's and the mother's genes into a Punnet Square, you find that the odds for a daughter would be: One in Two for each trait, giving you One in Four odds of having a normal daughter.

Normal: Comes from the definition of wild type which means the most common genes in a population. As most people do not have hemophilia, therefore, the mother does not have it.
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Postby envii » Wed Apr 06, 2005 11:37 pm

That is wonderful, thank you so much! :D
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Postby biostudent84 » Wed Apr 06, 2005 11:37 pm

Glad to be of service :)
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Postby HerpesCure » Tue May 31, 2011 9:38 am

What field did you choose after graduation?
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