Genetics solve the following genetics problem by determining...

Genetics: Solve the following genetics problem by determining how the following traits would assort. Use D/d for Rh. Remember Rh+ is dominant, and Rh- is recessive. Use A, B, and O for blood types. Remember it is a co-dominant system with O as the only recessive allele. The mother is type AB- and the father is type A+. List all possibilities for each parent's genotype using what you know from the information given so far (assuming you have only read (a) above). Now assume the father is heterozygous for both traits. What are the genotypes you will use for each parent? Create a double square (dihybrid cross) to assort these two traits. Show your work. What are the possible genotypes for each child? Provide a percentage (or a fraction) for each possible genotype. What are the possible phenotypes for each child? Provide a percentage (or a fraction) for each phenotype. What are the chances of having a child that is heterozygous for both traits? Provide a percentage and explain your answer. Is there any possibility that a child will be born to these parents that will show the recessive phenotype(s) for both traits? Explain: Why or why not?

Transcript

00:01 In order to answer this question, we have to talk about inheritance.

00:02 It says, solve the following genetic problems, my letter mining of how the following traits would assert.

00:07 Use d or rh, and remember that rh positive, it's dominant and rh negative is excessive.

00:13 And then it says you use a -b -o for blood types.

00:16 So if you have the homocygousocytes for a or the heterocygous because a and b are predominant and o is recessive, when you have these genotypes, you want to have type a -blot.

00:27 When you have the same but for b, it means homocygoyles for b or heterocygid, cycle for b, you're going to have type b blood.

00:34 When you have the heterocylus a and b, you're going to have type a .b.

00:38 And when you have the homocylusis b, you're going to have type o blood, okay? this is ab.

00:44 And for the rh group, when you have the homozygous dominant or the heterocylus, you're going to have rh positive, and when you have the homocylusiocylus, you're going to have rh negative.

00:55 So it says, and the mother is type a, b, negative.

01:01 So the mother is going to be a b, and she is homozygose, we say for the rh, a gene.

01:07 And it says, and the father is type a.

01:10 So the father is going to be homocygosegose dominant or heterocygose.

01:14 We don't know, so let's leave it like this.

01:16 Okay.

01:17 And a positive, so he's going to be also like this, because he can be homozygous dominoidicous, and up to here we don't know exactly his genotype.

01:26 So this is our cross.

01:28 It says, at least all possible, or possibility.

01:31 Is for each parent's genotype using what you know from the information given so far.

01:36 So according to this, this is the genotype for the mother.

01:39 Okay? we're 100 % sure that she, that this is her genotype.

01:44 And in case of the father, he can be, for example, dominant for all the genes, for the two genes.

01:50 He can be heterocygot for this gene and homocygousal dominant for this gene, or he can be homozygous dominant for this gene or he can be heterocygous for this gene.

01:57 Or he can be heterocygous for both genes.

02:00 Okay.

02:00 So these are the four possibilities.

02:01 For the father.

02:04 Now it says, now assume the father is heterocycles for both traits.

02:09 It means the father is like this.

02:11 It says, what are the genotypes you will see, you will use for each parent? and well, this is the genotype 4, the modern and this is the genotype 4, the father.

02:20 It says create a double square to assort these two traits.

02:25 So you have this proof.

02:31 Okay, you have to first find the gametes, and you're going to find the gametes by mixing each of this a and b with each of these d -ailils and the same here.

02:40 Each of this, oh, wait, i did a mistake here because the allils for this gene were not a in recessive a, were a, o, and b, okay? so i did a mistake here, and it says that the father is type a, so if he can be, let's delete all of this.

03:04 Initially, he was like this, okay? so if he is homocygous for both genes, he's going to be like this.

03:11 If he is heterozygoyles for this gene and homozygous dominant for this gene, you're going to have this.

03:17 If he is homozygous dominant for this gene and heterozygoy for this, you're going to have this.

03:21 And he can also be heterocygote for both genes.

03:24 So this is what you have the four possibilities.

03:27 So you just practically have to change the aggressive a that i placed here by the ovalid.

03:32 And well, in this case, the father genotype is going to be this one.

03:35 So you have here this one, okay? so the following gametes are going to be this, this, this, and this.

03:48 Basically, these are the gametes from this parent, and the gametes for this parent are going to be this one, this one, this one, and this one.

03:57 So they are asking us to make a panel square, so you have to place this part of these gametes in a panel square.

04:05 But actually here, as you can see here, this gamete is the same as this one, and this is a so you can just save time by placing only one.

04:15 This gametes only once.

04:16 Okay, so let's pick up a square.

04:17 You're going to have this and this.

04:20 At the end, the probabilities are going to be the same.

04:23 It is not going to affect anything if you place the four here or only two of them.

04:30 Okay? well, obviously because they are the same.

04:32 Okay.

04:32 If you place, if you decide to place the four limits here, then you're going to have two more goes and this is going to be the same at the end.

04:40 And when you find the probability for a specific genotype, it's going to be the same.

04:45 For example, if you place the four, that means, maybe for a genotype you're going to get maybe two quarters...

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