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Q: Genetics Problem ( Answered 5 out of 5 stars,   1 Comment )
Subject: Genetics Problem
Category: Science > Biology
Asked by: theaero-ga
List Price: $2.00
Posted: 01 May 2005 15:35 PDT
Expires: 31 May 2005 15:35 PDT
Question ID: 516603
Two lab mouse strains (A and B) differ in their DNA at 2 unlinked loci
(strain A has allele A at locus 1 and allele A? at locus 2; strain B
has allele B at locus 1 and allele B? at locus 2; no apparent
phenotypic difference is visible).

a. If strains A and B are crossed to give F1 hybrids, and 1 male and 1
female from the F1 are chosen to mate, what fraction of their progeny
(the F2) will be heterozygous for both genes?

b. If 1 male and 1 female are chosen from the F2 to mate, what is the
chance all their progeny will be homozygous at both genes?

c. How many mating pairs from the F2 (1 male and 1 female chosen at
random) should you establish to expect 1 pair to give all progeny with
BB at locus 1 and A?A? at locus 2?
Subject: Re: Genetics Problem
Answered By: neurogeek-ga on 02 May 2005 17:46 PDT
Rated:5 out of 5 stars
Considering the price you are willing to pay for your question, I will
assume that you are unable to access any genotyping service for your
mouse strains.  Although your mice don't have any phenotypic
difference, a genotyping service (such as that from Applied
Biosystems, linked below) can help you to keep the unpredictability of
breeding colonies to a minimum, especially in a sequenced animal like
the mouse.

Many larger universities also have in-house genotyping services, and
your animal care and use committee will probably want you to use that
route rather than establishing the number of mice you're talking about

Alright, on to the question.

a. The F1 generation is heterozygous for both genes (AB A'B').  F1
males and females can produce 4 different gametes for these loci.  For
a particular progeny to be heterozygous at both loci, it must have
been formed from complementary gametes: e.g. AB' and BA'.  There are
four such pairs, out of sixteen possible pairings.

b. For all of the progeny to be homozygous, the parents must be
homozygous and identical.  Complementary to the last question, for a
particular progeny to be homozygous at both loci, it must have been
formed from identical gametes: e.g. AB' and AB'.  There are four such
pairs, out of sixteen possible pairings.  The probability of picking
an identical mate for any particular F2 generation mouse is 1/16. 
Therefore the total probability is 4/16 * 1/16 = 4/256.

c. One pair needs to have the exact genotype BB A'A'.  One of every
sixteen in the F2 generation are BB A'A'.  So 1/16 for the first
parent, and 1/16 for the second parent.  (Assuming no sex-linkage to
these loci.)  Total you will need 256 animals, or 512 mating pairs, to
guarantee that one pair will always throw BB A'A' progeny.

In order to see this more clearly, I made a "Punnett Square" to
diagram the F2 cross with the four possible alleles in the F2
generation.  I used paper and pencil to draw mine, so can't show it
easily.  But, the following web site has a good explanation of Punnett
squares and some examples.

Punnett Squares
Explanation of a useful tool to answer this sort of question.

Applied Biosystems - a more practical approach to mouse genetics

Just for fun :-)
theaero-ga rated this answer:5 out of 5 stars
Thank you so much.

Subject: Re: Genetics Problem
From: politicalguru-ga on 01 May 2005 23:49 PDT
Thank you for your question.
I believe that to answer it well, your question will require more time
and effort than the average amount of time and effort associated with
this price. Here is a link to guidelines about pricing your question,
in the pricing guide:

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