Genes A, B, C, D, and E are linked on a chromosome and o...

Question

Genes A, B, C, D, and E are linked on a chromosome and occur in the order given.

Previous genetic linkage crosses have determined that recombination frequencies are 6% for genes A and B, 4% for genes B and C, 10% for genes C and D, and 11% for genes D and E. The sum of these frequencies between genes A and E is 31%. Why does the recombination distance between these genes as determined by adding the intervals between adjacent linked genes differ from the distance determined by the test cross?

Accepted Answer

Welcome back. Let's look at our next problem. It does. The frequency of recombination between genes A and B is 24%. And the following cross was done big A big B slash little A little B crossed with little, a little B slash little A little B. If 1000 progeny are produced from this cross, determine the number of progeny with big A big B slash little, a little B genotype. And our answer choices are A 0 B 190 c and D 760. Well, we know that uh being given the frequency of recombination, lets us figure out how many of the offspring are recombinant. Since recombination frequency is the number of recombinant offspring divided by the number of total offspring times 100%. So, what we need to do is look at the cross we're given and determine whether the genotype we're looking for is a parental genotype or a recombinant genotype. So our cross here, we have a heterozygous parent who has inherited the two dominant alls A and B from one parent, put that on top and then a line and the two recessive alls from the other parent. And that's crossed with a homozygous recessive parent. So we don't worry about crossing over with that parent since all of all are the same. Well, we could see right from looking at this, we are looking for the genotype in the offspring of big A big B over little a little B. But we could see right away that this is a parental genotype. Our offspring has inherited the two dominant alls together from the parent. So this is a parental genotype, not a recombinant genotype. But since we have our total number of progeny, we can say that our total number of progeny, 1000 minus the number of recombinant offspring will equal the number of parental type offspring. Now we have to remember to note that there are two possible parental genotypes. One that's inherited the dominant alls, one that's inherited the recessive alls. So the total number of parental type offspring has to be divided by two to determine the number of offspring with just that one particular genotype. So we know that the frequency of recombination is 24%. So the number of recombinant offspring will be 0.24, multiplied by 1000, the total number of offspring. And that will equal 240 recombinant offspring. So to find the number of parental type offspring, we're going to subtract Romney 1000 Total. And that will give us 760 parental type offspring. But again, we know there's two possible genotypes for parental type offspring. The one we're looking for and then as we said, the other one where they inherit the recessive all from the heterozygous parent. So we have 760 parental type offspring in total. So there will be of each genotype. So there are 380 offspring with the genotype we're looking for and that's choice. C 380 with the genotype big A big B slash little A little B. See you in the next video.

Key Concepts

Genetic Linkage

Recombination Frequency

Test Cross

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