1 you perform a standard dihybrid cross between two parents...

You perform a standard dihybrid cross between two parents with genotypes of: Parents: aabbddee and AABBDDEE (note there is no C gene) Answer the following questions: A. If you were to draw a Punnett square diagramming the progeny of a cross between two F1 siblings, how many "boxes" would it contain? B. How many types of gametes could be produced by each of the F1 individuals? C. What fraction of individuals would have the genotype: aaBbDDEe? D. What fraction of the progeny would be expected to have the dominant phenotypes associated with the A, B, and D genes and the recessive phenotype associated with the E gene? E. Assume the F1 individuals could mate with each other and you could observe the phenotypes of 1,000,000 of the progeny. About how many would display the recessive phenotypes of the A and B traits and dominant phenotypes of the D and E traits?

Transcript

00:01 So this question is really more math than anything else.

00:03 When we look at letter a, okay, we have a cross between a, b, d, d, d, e, and a, okay, we have four traits.

00:32 Okay.

00:33 So when looking at number one or a, if you were to do a punnet square, how many boxes will you you have? well, you have four traits.

00:46 And each trait, if you foil it out, one, two, three, if you foil it out, it's going to be four to the fourth, and that's going to be 256.

01:12 It would be insane, right? let's look at b.

01:17 How many types of gametes can produce by each of them? well, you have four traits, each with two alleles.

01:28 So it's going to be 16.

01:30 Gametes.

01:38 What fraction of individuals would have the genotype? so the genotype is little a, little a, big b little b big d big d big e little e in that cross zero so the possibility of having this when you have that and that is none okay sorry this is a cross between the okay so we're crossing this we're crossing this so we're crossed these two these were our parents and it gave us our f1 generation right here and then we're looking at a cross between those so the chance of getting this is one out of four right because a cross between that and that one out of four a chance of that is one half right because when you put those two together it's a one -half chance you'll get that here cross between those it's one -fourth and this cross between that is one -half which is going to equal 1 out of 64.

03:28 Okay.

03:30 And what fraction of progeny expected to have dominant phenotypes with a, b, and d, and the recessive phenotypes associated with e? so again, we're looking at you have a three -force for a to have a three -force for a to have a.

03:55 A dominant phenotype of a, right? because when you look at this cross, it's a three -fourth chance that you're going to get that or that.

04:07 And three -fourth chance you're going to get this.

04:13 And three -fourth chance are going to get this.

04:16 A dominant phenotype over here.

04:18 And then it's a one -fourth chance that you're going to get a recessive phenotype there...

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