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I can not figure out this Genetics problem? Please help?


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Genes for color blindness and hemophilia are linked on the X-chromosome in humans with a recombination frequency of 30%. A woman whose father was a color blind hemophiliac and mother was normal with no history of color blindness and hemophilia wants to have children. The prospective father is normal with no history of either colorblindness or hemophilia in his family. what is the probability that she will have a normal child?

1.) 0.70

2.) 0.675

3.) 0.50

4.) 0.35

5.) 0.30

6.) 0.15

I have to turn it in by 5:45

any help is greatly appreciated
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Is this how you do it and is this the correct answer?


Since man has only one X chromosome, she must have one X defective and one X normal (from mother).
In 30% of cell divisions the chromosomes recombine and form two defective X chromosomes (each for other disease, but they ask for normal kid, so any of these chromosomes will be ruled out). In the other 70% cases the chromosome will be defective in 50% and normal in 50%, i.e. 35% of total cases. Thus 4.) is the right answer.

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An easier way to work through these is knowing the distributions. For 30% recombination it'll be .35-.15-.15-.35. But that's only for the heterozygous parental gametes. Since you know all girls will be normal since the father is normal that is at least 50%. So the earlier distribution applies only to the males. So you have a probability of .5 for a girl who are always normal, P(g). But a probability of .35 for normal,P(n), and .5 for a boy,P(b). Since they're independent P(n*b)=.5*.35=.175. So P(nb OR g)=.5+.175=.675

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  • 2 weeks later...

hmm. I don't get a value which is the same as any of the answers. I will post my line of thinking anyway.


In order to understand what happens, you need to consider the outcome of recombination and no recombination with respect to meiosis. Before that though we can already tell that any girl will be normal because the father carries a normal X chromosome. So 50% at least will be normal.


Before calculating the probability of a male being normal or afflicted with either genetic disease, it's best to think through meiosis first.



Meiosis will only be considered for the mother since she is the one carrying her father's defective X chromosome. In meiosis 1 there is a bivalent of her maternal and paternal chromosomes, where both of the paternal chromatids are defective and none of the maternal chromatids are defective.


(70% chance) No recombination scenario: the maternal chromatids remain normal and paternal chromatids remain fully fective (with both disease alleles). Two chromatids have no defective alleles, and the other two chromatids have defective alleles. 2 gametes have no defective allele, 2 gametes do. 50% chance of being normal.


(30% chance) Recombination: One maternal chromatid will recombine with a paternal chromatid, the result of which is both chromatids become a carrier of a defective allele due to recombination unlinking the disease alleles. Other maternal chromatid remains normal and the other paternal chromatid remains defective. The end result is three gametes carry defective alleles and one gamete is normal, 75% are defective allele carriers and 25% are normal.


The calculations (only for the probability of a normal outcome):


No recombination scenario: 0.7 x 0.5 = 0.35


Recombination: 0.3 x 0.25 = 0.075


add them together for overall probability of having a normal male = 0.35 + 0.075 = 0.425 or 42.5%


42.5% only considers probability for a male. equal chance to have a male or female so 42.5/2 = 21.25%


50% (chance of female being normal) + 21.25% = 71.25% chance of a normal child. I go with option 1.


It seems I am wrong. I disagree with previous posters' calculations since recombination event results in 3 out of 4 gametes being defective allele carriers. .3 x .25

Edited by jp255
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