In the rats identified in Problem 10, a third independently assorting gene involved in the determination of coat color is the C gene. At this locus, the genotype C– permits expression of pigment from genes B and D. The cc genotype, however, prevents expression of coat color and results in albino rats. For each of the following crosses, determine the expected phenotype ratio of progeny.
bbDDCc×BBddCc

Question

In the rats identified in Problem 10, a third independently assorting gene involved in the determination of coat color is the C gene. At this locus, the genotype C– permits expression of pigment from genes B and D. The cc genotype, however, prevents expression of coat color and results in albino rats. For each of the following crosses, determine the expected phenotype ratio of progeny.
bbDDCc×BBddCc

bbDDCc×BBddCc

Accepted Answer

Hi, everyone. Welcome back. Let's look at our next problem. It says for an independently assorting gene, short hair. Big S is dominant over long hair, little S in rabbits and black skin. Big B is dominant over brown skin, little bee. When the male and female with the heterozygous genotype, Big B Little B big S little S from the F one generation are self crossed. What is the probability of having the genotype? Big B big B big S big S in the F two generation? So we're looking for that genotype with homozygous dominant in both traits. We have two parents who are heterozygous and both traits being crossed. So we have Big B Little B big s little s crossed with Big B little B, big S little S. This is a die hybrid cross. Now to find out this probability, we could make a Big 16 square punt square. But there's a simpler method for calculating this probability, which is we know that the probability of two events happening is the product of the probability of each of the events happening because our genes independently as so we can look at the probabilities separately for each one. So we could say, what's the probability of having the genotype? Big B, Big B? Well, now we're just looking at a mono hybrid cross. So let's look at Gene B first separately. So if you've got a mono hybrid cross, Big B Little B cross with Big B Little B, well, that's a very simple genotypic ratio of 1 to 2 to one. The ones being the homozygous homozygous dominate and homozygous recessive and the two being heterozygous. So we're looking for Homozygous dominant. So that probability will be one fourth. So that's the probability of being homozygous dominant for gene B. Now we'll multiply that by the probability of being homozygous dominant for gene S, well, gene S is going to be the very same scenario, parents are heterozygous big S Little S cross with big S little S, that's the 1 to 2 to one ratio. So the probability of being homozygous dominant is 1/4. So our probability of both things happening being homozygous dominant for both genes is the product of both probabilities. So 1/4 multiplied by 1/4, which gives us 1/16. So when we look over at our answer, choices, we have choice A 9/16 choice B 3/ choice C 1/16 or choice D, none of these. So our answer is right here in choice C 1/ as the probability of having the genotype, Big B big B big S big S in the app two generation. See you in the next video.

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Key Concepts

Independent Assortment

Genotype and Phenotype

Punnett Square