The genotype r / r ; p / p gives fowl a single comb, R /-; P...

The genotype $r / r ; p / p$ gives fowl a single comb, $R /-; P /-$ gives a walnut comb, $r / r ; P /-$ gives a pea comb, and $R /-; p / p$ gives a rose comb (see the illustrations) Assume independent assortment. a. What comb types will appear in the $\mathrm{F}_{1}$ and in the $\mathrm{F}_{2}$ and in what proportions if single-combed birds are crossed with birds of a true-breeding walnut strain? b. What are the genotypes of the parents in a walnut $x$ rose mating from which the progeny are $\frac{3}{8}$ rose, $\frac{3}{8}$ walnut, $\frac{1}{8}$ pea, and $\frac{1}{8}$ single? c. What are the genotypes of the parents in a walnut $x$ rose mating from which all the progeny are walnut? d. How many genotypes produce a walnut phenotype? Write them out.

Transcript

00:01 In order to answer this question, let's talk about inheritance.

00:03 It says the genotype homocygose decisive for r and p calls for a single comp.

00:13 Then you have that the homozygose, or well, the dominant r and the dominant p are going to call for walnut.

00:23 Then you have that the homozygote recessive for r and the dominant p gives p comp.

00:32 Then you have that the dominant r and dachesis p calls for goes comp like this okay and it says assume independent assortment it's just what what comp types will appear in the f1 and in the f2 and in what proportions if single comp a single comp birds it means with this genotype are crossed with birds of a bird's true breathing walnut strain.

01:09 So true breathing means homozygos.

01:12 So in this case that true breathing or homozygos, a walnut, walnut camp birds are going to be like this.

01:24 So remember that a person or well a parent transmits one of their alleles for each gene.

01:32 In this case, for example, focus only on the argyne.

01:35 This parent has only recessive alleles to give to the offspring and this parent only has dominant other links to give to the off -pring so one dominant r is going to come from this parent and one -gishifar is going to come from this parent so you have that both parents are hitterocytocles like this okay or i'm sorry both all of the off -pring are going to be hit hit hit -eot cycles this is the only possibility for the f1 and this f1 calls for as you have the dominant r and the dominant p there are all the f1 are going to have walnut comp now in order to produce your f2 you have the following in this case, you have to make a dichary panel in square and you're going to do that by mixing.

02:14 First, you have to make the gametes to get the gametes and you're going to get the gametes by mixing each of these are al -a -lil with each of these p -allel and as both panels have the same genotype, both parents are going to have the same gametes, this, this, this, this, and this, and the same in this case.

02:30 Okay, homo saigon dominant, homo saigos -dominal, hithro -sai -go's, hitor -sai -go -g, heteros, home -sai -go's, hito -sign, heteros, h estate goes, hector -sig ombian, h40, heteros, hadoz, gitro -diminate, hitro -discence, h jğimiz, heteros, h الج, g confusion, heterocylus, heterocyclos, homocyclos, homo -rescentive, homo -sacizant, homo -sacizisive, heterocygote, heterocygote, heterocygots, heterocyg, heterocyg, homo -sigisive, homo -sacos, excessive, heteroseph, heteroseph, and a double -phomocylus, recessive.

03:33 So, in this case, they are asking you for the type of combs and the proportions, okay? so in this case, first let's look for those that have that single comp.

03:48 You have that out of 16 possibilities, only this one is going to have single comp.

03:53 So you have one for single comp.

03:56 Now for walnut, for walnut comp, you need dominant r and dominant p.

04:01 So you have in this case, 1, 2, 3, 4, 5, 6, 7, 8, 9.

04:14 I'm sorry, this is not because this is dominant and excessive.

04:17 Okay, so in this case, let's change that.

04:21 4, 6, 7, 8, and 9.

04:24 So you have 916s for walnut.

04:30 Then for p comp, you need an excessive r in the dominant p.

04:34 So you have 1, 2, and 3.

04:40 316s for p comp.

04:46 And finally, you have 1, 2, and 3.

04:52 For dominant r and b6p, that is for, for goes come.

04:57 So you have 316s for goes count.

05:00 So these are the proportions and the phenotypes for each of these cases.

05:04 Question b says, what are the genotypes of the parent signal walnut with the ghost meeting? so walnut is like this, and the ghost is like, like this, dominant r and a specific p.

05:20 Okay...

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