In rabbits the dominant allele b causes black fur and the...

In rabbits, the dominant allele "B" causes black fur and the recessive allele "b" causes brown fur. For an independently assorting gene, the dominant allele "S" causes short fur and the recessive allele "s" causes long fur. A heterozygous black rabbit with short fur (heterozygous for both genes) is crossed to a homozygous (for both genes) black rabbit with long fur. (A) What are the genotypes of the parents? (B) Using a single Punnett square, determine the proportion of the F1 offspring that will have long, black fur. (C) Using a branch diagram, determine the proportion of F1 offspring that will have short, black fur. (D) Is it possible to get brown rabbits in this cross? If yes, in what proportion? If not, explain why.

Transcript

00:01 Okay, i'm going to answer this question.

00:02 We'll have to talk about inheritance.

00:03 It says in gravity, the dominant allele possess black fever.

00:08 So this is the black fever allele.

00:13 And the small b is the brown fever.

00:17 Brown field.

00:18 So we can have practically homocybo -sig dominant or heterocygose.

00:24 And the phenotype that is going to be coded by this genotype is a is black fever.

00:31 And when we have the homocybo -sacios excessive, we're going to to have brown fur.

00:37 It says for an independently assorting gene the dominant allele is so we have a big s causes short fear so this is short fear, short fear allele, and the small s causes the long fur.

00:54 This long fuel allele.

00:58 Okay, it says a heterocygous black rabbit with short fear, it says it is heterocygous for both the genes, it means that you know that it becomes to be.

01:10 This.

01:11 Okay, it's close to a homocygous for both genes, black rabbit with long fur.

01:18 It means black rabbit that is homocygose and long fear, homocygose excessive.

01:24 So you have this.

01:29 Okay, so this is the genotype for over two gavits.

01:33 Question a says, whether are the genotypes of the parents, well, these are the genotypes of fathers, of the parents, okay? so this is a question a, question vices using a panel square that will have a proportion of the f1 off print that will have a long black fewer well they are asking us to use a panel square so we have to make a diet hybrid panel square well a smaller panel square practically okay remember that in order to get a dachamids from each parent for two genes we have to cross this big b with this bigs and then this big v with this small s and then the same we're going to do with this small b allele so we're going to have a following possibilities b s b s b s and small b s and the same we're going to do here and we're going to have only this because we're going to well this is the only possibility that we're going to get here okay we're actually get this this and this but all of them are the same so you can resume them as one.

02:41 Now we're going to make a panel square that is going to be.

02:44 We're going to place at one side, this gametes, and at the other side, this gametes.

02:48 So you're going to get this, this, and finally, this, okay? and you're going to cross with the only possibility for this part.

03:09 And you're going to get the following possibilities.

03:20 And finally, ethereal cycle for b and homozaiw can see for s.

03:23 So these are the possibilities.

03:24 They are asking us determine the proportion, it means a fraction of the f1 off -print that will have long fever and black fur.

03:34 So you have one, two, three, and four possibilities.

03:37 You have a total of four possibilities.

03:39 All of those four, how many of them are going to have black fever? well, first, black fever.

03:45 It means we're looking for either this or this.

03:48 And long fear, it means with this almost i go.