- November 11, 2011 at 5:30 am #15670
3. In humans, hemophilia and red/green colorblindness are both due to sex-linked recessive alleles. The dominant alleles produce normal blood and normal vision. A hemophilia, colorblind male marries a non-hemophiliac, normal-visioned female, whose father had hemophilia and colorblindness. What kinds of offspring would they be expected to produce and in what ratios?
how do you do a dihybrid cross with two sex traits? thanks
- November 11, 2011 at 7:31 am #108016aptitudeParticipant
All you have to do is use the laws of probability to determine the ratios by multiplying the individual probabilities together.
1. The male must carry the recessive colorblindness and hemophilia allele because he is color-blind and has hemophilia
2. Since the female’s father was a hemophiliac and colorblind, he must have passed on these two recessive alleles to his daughter. Since the daughter is neither colorblind nor a hemophiliac, she must be heterozygous at both loci.
Set up two punnet squares, and figure out the probabilities. Then simply multiply them together.
- November 11, 2011 at 8:34 am #108020
So are you saying that I do one cross for the hemophilia trait and one for colorblindness?
- November 11, 2011 at 5:57 pm #108050aptitudeParticipant
- November 12, 2011 at 2:20 am #108057monstermonster456Participant
If you wanted to do the dihybrid cross, it would be X^hcY x X^HCX^hc
- November 13, 2011 at 3:15 am #108072
so the father is recessive for both traits X^hc Y and the mother is a carrier hetero X^HC X^hc.
so from X^hcY x X^HCX^hc, how do i break it down and set up the cross?
- November 15, 2011 at 5:47 pm #108125zombiesaganParticipant
If you were to trying to determine them individually (which is much easier), than you have to think of what each parent has to offer: for example, in colorblindness, the father can either give X^c or Y and the mother can give X^C or X^c. Those children who wind up XY will be male and those with XX will be female and if they have at least one C they will not be colorblind (note that a male only needs one recessive gene to show the trait because there is nothing on the Y).
- November 16, 2011 at 2:40 pm #108157ChesneMDParticipantcode :
X^CH| X^CH X^ch | X^CH Y
X^ch| X^ch X^ch | X^ch Y
That is, assuming I got the information correctly, which in the first post said a carrier mother, ch father.
So the genotype ratio would be 1:1:1:1, and the phenotype is 1:1:1:1 (because it’s male an female), I think. I honestly cannot recall how to do the ratio part for sex linked squares at the moment.
- November 16, 2011 at 10:01 pm #108186CatParticipantquote Timmai:
Break down information to points of facts:
1. Both conditions sex linked => on X chromosome
2. Both recessive => pronounced in males as they are hemizygous for x chromosome.
Thus, “a non-hemophiliac, normal-visioned female, whose father had hemophilia and colorblindness” means that she is heterozygous, Xx, where x caries both hemophilia and colorblindness inherited from her father.
3. “A hemophilia, colorblind male” can ONLY mean that his genotype is xY where x caries recessives for both conditions.
4. Now you can do a cross of xY (“A hemophilia, colorblind male”) and Xx (“a non-hemophiliac, normal-visioned female, whose father had hemophilia and colorblindness”).
When you do this cross, you will find that 1) in this case two alleles are linked and, therefore, 2) ratios of this cross will depend on how closely they are linked.
5. If the alleles are closely linked and recombination frequency = 0, you do a simple cross of xY x Xx and get 50% normal and 50% hemophilia + colorblind.
6. If alleles are far apart, than you have recombination in the heterozygous female:
X=(H,C) ,x=(h,c) can produce gametes paternal (H,C), (h,c) and recombined (H,c), (h,C)
While male will have ONLY x=(h,c).
Now, all male progeny will have one x chromosome from the mother. =>
(H,C) Y a non-hemophiliac, normal-visioned
(h,c) Y hemophilia, colorblind
(H,c) Y non-hemophiliac, colorblind
(h,C) Y hemophilia, normal-visioned
While females will have an x chromosome from the mother and SAME x chromosome from the father (as only one possibility is available). =>
(H,C) (h,c). a non-hemophiliac, normal-visioned
(h,c) (h,c). hemophilia, colorblind
(H,c) (h,c). non-hemophiliac, colorblind
(h,C) (h,c). hemophilia, normal-visioned
Unless you know recombination frequency, you cannot know ratios produced by this cross.
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