A pure-breeding fruit fly with the recessive mutation cut wing, caused by the homozygous cc genotype, is crossed to a pure-breeding fly with normal wings, genotype CC. Their F₁ progeny all have normal wings. F₁ flies are crossed, and the F₂ progeny have a 3:1 ratio of normal wing to cut wing. One male F₂fly with normal wings is selected at random and mated to an F₂ female with normal wings. Using all possible genotypes of the F₂ flies selected for this cross, list all possible crosses between the two flies involved in this mating, and determine the probability of each possible outcome.

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Welcome back. Here's our next problem. It says four different types of p plants are self fertilized to obtain the following three groups of progeny. Got group, one round seated pea plants, 101 wrinkled seeded pea plants. 99 group two round CDP plants, 99 wrinkled seeded pea plants, pea plants, zero, group three round seated pea plants, 90 wrinkled seeded pea plants. 30 identify the group of progeny. That is the result of a cross between two hetero zegas parents. So let's think through this here. We have 200, I guess parents as the cross. So we'll just put up a simple punnett square here. So it's just a mono hybrid cross. So just four boxes. So across the top. Big R little R, we know we have two hetero parents. So on the left also big R, little R and we know that our four boxes must be top left, big R, big R below it, big R, little R on the right side, big R, little R on the top and little R little R on the bottom. So we see our phenotype is because our questions ask us about the phenotype of the offspring should be in a ratio of 3-1 round, too wrinkled. And of course, at this point, we should mention if you didn't remember off the top of your head, which shape was dominant. When you look at the groups they give you, we see at least one group here where you had all the plants being around, seated and zero being wrinkled. So that tells us right away that round must be the dominant phenotype and wrinkled, the recessive. So if you need that tip, um that's there and the groups that you're given. So we've got here 321 round to wrinkled. When we look at our groups, we see that group one shows basically a ratio of 1 to 1, 101 round 2 99 wrinkled. So that we wouldn't expect to see unless we had a hetero cygnus parent crossed with a homesickness recessive. That's where we see that 50 50 ratio. So group one is not correct. So we can eliminate choice A which is group one only And we can eliminate choice d which is group one and 3 only. So now let's look at group to group two is the one that has all of the offspring being round, seated, zero of them being wrinkled. So that's that offspring of or a ratio of 1 to 0. If all of the offspring around, we'd expect to see that would be the, the result of a cross where you'd have homocide is dominant, cross with homesickness, recessive, all the offspring would be hetero sickness and therefore all would show the dominant trait. So here we are with, so group two, we know is not correct. So we can eliminate choice B group to only. Now, let's move on to group three. We see that has round plants, 90, wrinkled, 30. So that is that ratio of 3-1. Oops, we'll go ahead and put that not in red since that's correct, that's our ratio that we want 31. So I'm going to write that in blue. So we know that choice C Group three only must be our answer. And as I note here, I drew out the punnett square, but it can be useful to remember with a mono hybrid cross. The different ratios. If you've got two hetero side gets in a mono hybrid cross, your fanatic ratio will be 3 to 1 dominant to recessive phenotype. So that's something that's good often to just keep in your head so that you don't have to sit here and make a punnett square on a test for every question like this. So again, which of the groups of progeny resulted across between two hetero parents. And that's the one that showed this 3 to 1 ratio. Choice C group three only. See you in the next video.

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Key Concepts

Mendelian Genetics

Genotype and Phenotype

Punnett Square