Two genes are located on the same chromosome and are known to be 12 mu apart. An
ID: 8905 • Letter: T
Question
Two genes are located on the same chromosome and are known to be 12 mu apart. An AABB individual was crossed to an aabb individual to produce AaBb offspring. The AaBb offspring were then crossed to aabb individuals.A. If this cross produces 1,000 offspring, what are the predicted numbers of offspring with each of the four genotypes: AaBb, Aabb, aaBb, and aabb?
B. What would be the predicted numbers of offspring with these four genotypes if the parental generation had been AAbb and aaBB instead of AABB and aabb?
Explanation / Answer
Hi, so we start with figuring out whats being crossed. We know for part A that we're crossing AaBb with aabb. Lets first see what gametes can be produced from the AaBb individual. By the rules of independant assortment, we know it can produce gametes with AB, Ab, aB, and ab if each of the A alleles match with each of the B alleles. Similarly, the aabb individual can produce only ab gametes since it only has a little 'a' and a little 'b'. Now we set up our 4x4 punnett square ab ab ab ab AB AaBb AaBb AaBb AaBb Ab Aabb Aabb Aabb Aabb aB aaBb aaBb aaBb aaBb ab aabb aabb aabb aabb So we see that 25% of the time we get AaBb, 25% Aabb, 25% aaBb, and 25% aabb, an even split. Out of 1000 offspring, we'd expect 250 of each. B) Had the original parental generation been AAbb and aaBB, we cross them to get our "first generation". Note again that AAbb can only make four Ab gametes, and the aaBB individuals can only make aB gametes. Crossing, Ab Ab Ab Ab aB AaBb AaBb AaBb AaBb aB AaBb AaBb AaBb AaBb aB AaBb AaBb AaBb AaBb aB AaBb AaBb AaBb AaBb So you see we have the same heterozygous genotypes. Therefore, when we procede again as in part A, we would get exactly the same results! So it didnt matter in the end. Hope that helps! Jonathan
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