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.

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.

Key Concepts

Genotype-Phenotype Mapping

Genotype-Phenotype Mapping is the process of correlating the genetic makeup (genotype) of an organism with its observable traits (phenotype). This mapping is critical in studies that involve determining which combinations of alleles result in specific physical characteristics, such as the various comb forms in fowl.

Dihybrid Cross

A Dihybrid Cross examines the inheritance of two distinct traits simultaneously by crossing individuals that are heterozygous at both loci. It uses Punnett squares and probability rules to predict the ratios of genotypes and phenotypes among the offspring, which is necessary for determining the outcomes of crosses involving genes affecting comb type.

Epistasis and Gene Interaction

Epistasis occurs when the effect of one gene is modified by one or several other genes. In the context of the comb types in fowl, interactions between different genes (such as those determining comb form) can produce unique phenotypes that are not simply the sum of each individual gene's effect, and understanding these interactions is key to interpreting the observed inheritance patterns.

Independent Assortment

Independent Assortment is the principle that alleles of different genes separate independently of each other during the formation of gametes. This is crucial in dihybrid crosses where two or more traits are studied simultaneously, leading to a variety of genotypic and phenotypic combinations in the offspring.

Mendelian Inheritance

Mendelian Inheritance is the principle by which traits are passed from parents to offspring through discrete units known as genes. It involves understanding concepts such as dominant and recessive alleles, which determine the observable characteristics or phenotypes in organisms, as well as how these alleles segregate and assort independently during gamete formation.

Allelic Dominance

Allelic Dominance refers to the relationship between alleles, where the dominant allele masks the expression of a recessive allele in a heterozygous genotype. This concept is essential for understanding how different combinations of alleles at a locus result in distinct phenotypes, such as the different comb types observed in fowl.

Transcript

00:02 In chickens, comb is inherited in this pattern.

00:05 So if we know that little r, little r p, p yields a single comb, and big r something, big p something yields a walnut comb, i'm going to underline my peas that are recessive because you can't tell the difference between the capital p and a lowercase p.

00:32 Little r little r big p something yields a pea comb and a big r something two little peas yields a rose comb.

00:43 Now we have enough information to answer these four questions.

00:49 So let's start with a.

00:51 A is a situation where a single comb chicken was bred with a true breeding walnut comb.

00:59 And so walnut has to be big r, big p.

01:04 But since it's true breeding, we know it's homozygous for both, dominant for both traits.

01:11 Single is a double recessive.

01:14 And so we can easily figure out the parents.

01:17 That means the f1 is a heterozygote.

01:22 Oops, i forgot to underline those.

01:24 And the f2, because these are independently assorting genes, yields a 9331 ratio.

01:30 With big r something big p something yielding walnut comb and then two little r's big p something is a p comb big r something two little p's will be a rose comb and of course the double recessive is going to be the single comb so that takes care of a let's take a look at b b.

02:01 B says what happens when you read a single, i'm sorry, a walnut comb with a chicken who has a rose comb.

02:13 Well, remember, walnut is big r something, big pea something.

02:18 And rose is big r something two little peas.

02:24 And that gives us quite a lot of information to work with.

02:28 And so we also know that the f1 is three eighths.

02:33 So again, let's write down what we know.

02:38 And three -eighths are walnut, one -eighth p, and one -eighth single.

03:00 Okay, so we can figure this out here.

03:03 Because first, let's look at these genes separately.

03:07 That'll make our life easier.

03:09 If you look at the r's, there is a three -to -one.

03:16 Ratio of ours.

03:23 The only way to get a three to one ratio for ours is to make sure that the parents were heterozygous, right? so if we did a quick cross of these, we would see that you would get three dominant phenotypes and a single recessive.

03:44 So the only way to get the three to one ratio here is for us to have homozygous r parents.

03:53 So we can fill those in...

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