Few Genetic Questions!

[Igal0s]

1. When dealing with a dihybrid cross, say I am told that two genes are 22cM apart and that a cross of oossXOoSs (testcross) was done. In such cross, if I am right, I should predict an equal, 25% proportion of the 4 different phenotypes. My question is, If i am asked what number, out of the 100 from the cross will exhibit a particular phenotype (aka 1 of the four, 25%), how do I account for the 22cM distance of the two genes (O for round/oval shape and S for red/yellow)?

2. What is the probability that an autosomal dominant condition will be inherited to the child from two affected parents? Please show workings. (I got 3/4?)

3. I have 4 gene loci on a single chromosome, A,B,C,D. Looks like this:
A <-15mu-> B <-5mu-> C <-40mu-> D
———————————————————

The question states, using this information, an organism hetrozygous for A and B alleles arranged in trans would produce what pair of gametes in low frequency?

A.aB and Ab
B.AB and aB
C.AB and ab
D.aB and ab

Thanks heaps in advance!

[mith]

1. 22cM means that 22% of your offspring will be recombinants.
2. I think it’s larger. Consider the different genotypes the parents can be.

[Igal0s]

I understand but not all 22% will produce a genotype that does not lead to the required phenotype. The likely answers (its a MPQ) to me seems either 25 or 22 (with allowing some room for crossing over) but i am highly dubious of this answer as it seems they placed it there for people who have no clue and will read figures of the question (22%).

With Question 2, you reckon it’s even higher than 3/4? The way I see it, the only combination that will produce a phenotypically healthy child is hh. Hence, assuming its a HhxHh cross, it is 1-1/4. But I am really unsure if I can utilize such a assumption so simply? 😕

Thanks in advance again!

[mith]

1. A correction to my previous post, it should say that the gametes are recombinants.

2. You can also have a HH X HH cross — which produces no healthy children.

[biomaster01]

for your first question it means that because the genes are only spaced by 22 map units then the genes are said to be linked and therefore there will be higher number of offspring that will show the parental phenotype and not the recombinat one!! therefore instead of having a ratio of 1:1:1:1, two of the types of offspring will be much higher percentage of the total offspring and two will be far less and the one you want will be roughly 11out of the 100. hope that helps!

[Author]

Posts