Determinig linkage map and parental genotype

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sportboy4life13
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Determinig linkage map and parental genotype

Post by sportboy4life13 » Fri Jul 03, 2009 7:36 pm

I am currently in a genetics class and am having a difficult time with this sections homework. I've looked through the chapters and still don't understand how to figure out the problem. Any help would be greatly appreciated.

2. Female animals heterozygous for three recessive alleles (a/a+, b/+ , and c/c+ ) are mated with normal males. The phenotypes of the F1 progeny are as follows:

females:

a+b+c+ 1991

males:

a+b+c+ 31

a+b+c 33

a+bc+ 443

a+bc 2

ab+c+ 1

ab+c 427

abc+ 25

abc 38

Explain the results. Show the genotypes of the parents, and construct a linkage map showing the correct order and distance of the loci.

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Jesse2504
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Post by Jesse2504 » Sat Jul 04, 2009 9:03 am

Your females are heterozygous for 3 recessive alleles, so they will have the genotype: AaBbCc

aabbcc is homozygous recessive
AABBCC is homozygous dominant
AaBBCc is a mix, heterozygous for Aa and Cc alleles, homozygous for B.

When you say your males are "normal" does this imply they are AABBCC?

How come your offspring only have 3 alleles instead of 6?
They should look something like:
AaBBCc
AAbbCc
AabbCc
Jesse

I spit in the mouth of a god, who whispers in the minds of the children

"The most incomprehensible thing about the universe is that it is at all comprehensible" - Albert Einstein.

littlemonkey
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Post by littlemonkey » Sat Jul 18, 2009 5:04 pm

I'm sorry about my English. I have just practiced using it.
In this problem, we have trouble with the ratio between male and female (1:2). This ratio is specialized for dead alleles. We can notice that there are enough phenotypes in male. In this case, may be there is a mistake of the number (may be 991 female). I can give another hypothesis is that there is another allele in female's genotype. The male who is homozygous or hemi-homozygous of this alleles will be dead.
There is a difference between male and female, it proves that these alleles are sex-linkage. And the ratio of males proves that three alleles (and other in my hypothesis) are located in X-chromosome.
Now, we consider the ratio of male's phenotype:
The parental class (a+cb+ plus ac+b) is 87%. The double crossovers (ac+b+ plus a+cb) are 0.3%. It proves that the central allele is c.
The single region 1 ( between a and c) crossovers are 5.8%. The single region 2 (between b and c) crossovers are 7.1%.
Therefore, the region1 map distance is : 5.8 + 0.3=6.1 cM.
The region 2 map distance is: 7.1 + 0.3 = 7.4 cM.
Hence, the distance between a and b is 6.1+7.4 = 13.5 cM.
Now, we try to explain why there is a difference between male and female ( in number of offspring - ratio 1:2). The female has a heterozygous genotype, Bb for example, and it is a nonsex-linkage alleles. and the male has genotype, bb. When they form F1, the males, who receive alleles B will die - Bb. The males have genotype bb is normal. So, the number of male is one half of number of female. (This result may be the form of the epitasis between the gene and cytosol)
I know something is wrong in my hypothesis so that please help me to fix it. Thanks!

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