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Ch. 14 - Mendel and the Gene

Chapter 14, Problem 9

In parakeets, two autosomal genes that are located on different chromosomes control the production of feather pigment. Gene B codes for an enzyme that is required for the synthesis of a blue pigment, and gene Y codes for an enzyme required for the synthesis of a yellow pigment. Green results from a mixture of yellow and blue pigments, and recessive mutations that prevent production of either pigment are known for both genes. Suppose that a breeder has two green parakeets and mates them. The offspring are green, blue, yellow, and albino (unpigmented). Based on this observation, what are the genotypes of the green parents? What genotypes produce each color in the offspring? What fraction of the progeny should exhibit each type of color?

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Welcome back. Let's look at our next problem here. It says two sets of genes on different chromosomes are responsible for the coat color and animal X. Gene be coats for the brown color and gene G for gray. Beige colored coat results when these genes are in hetero ziggy's condition and in case both the genes are recessive, the coat color is white. Suppose a male of X is made with a female of X. Both having a beige coat. What fraction of the progeny do you expect to have white coat color? Okay, that is a lot of information. A lot of words coming at us um will kind of highlight some things to make them clear. So we've got um one thing I think came straight is the beige colored coat results from hetero sickness genotype. And then in red here, I'm going to highlight that if both genes are recessive, we have a white coat color. So let's keep that straight. And we're thinking about parents and progeny and we've got two jeans, I'm gonna drop them down. We have jean B for braun and gene G for gray. And we know that in both cases when the jeans are both recessive, the coat color is white. So that's going to lead us to say that Big B will be brown since we know the gene for brown color and Little B will be white. Big G. Will be gray and little G white. We know that the genotype of Little B. Little B. Little G. Little G is white. I'm gonna highlight that in red to match it with the information we were given in the problem. And we know that our cross, we say that both the male and female have a beige coat well beige means hetero Ziggo. So the cross we're looking at is parents that are hetero ziggy's for both traits. I'm gonna highlight that in blue. We know they're beige. So kind of again matching that up with information we're given in the problem. Now when we have two hetero six parents being made into each other and to jeans, this is what's known as a dye hybrid cross. And it's kind of useful to remember that when you have a di hybrid cross your offspring's phenotype. So in this case the coat colors will be in a ratio of 9 to 3 to 3 to 1. Obviously that's out of a total of 16. And that will be the phenotype of showing the um both dominant traits In three and 3 showing one dominant and one recessive trait And one showing both recessive traits. So in this case one will have the little B. Little B. Little g. Little G genotype which gives us the white phenotype which is what we're looking for that fraction that has the white coat color. So that's going to be one out of 16. So we can go ahead and say that our answer will be choice A. Here 1/16 of the progeny we'll have this white coat color. But if you don't remember this ratio, that's again, kind of a useful thing to have in your head? But you might not remember it. Um We're just gonna go ahead and look quickly at a punnett square from this meeting, drink this down a bit and again, we know here's our parents. So we know that our gametes that are possible from these parents are big be big G. Big G, little G, little B, big G. And little B little G. Both parents have the same genotype. So it'll be the same gametes on both sides of this. And again, we're looking for progeny that have all recessive genes. So we need where they inherit the recessive genes from both parents. And that's only going to be this square right here. That's the only one that will have the combination little B, little B, little g, little G. Because all the other squares, we don't even need to write it all out. We see anything down here, We'll have at least one dominant trait. Same with this row. Same with this room. This big B. Will be in all of these offspring. This big B will carry across this entire row. This big G across this entire row. And this big G. Down this entire row. Just one square out of the 16 square punnett square will have the both recessive genes that give us the white coat color. So again, that gives us that answer. Of 1/16 choice A. For what fraction of the progeny would you expect to have the white coat color? See you in the next video.
Related Practice
Textbook Question

In garden peas, yellow seeds (Y) are dominant to green seeds (y), and inflated pods (I) are dominant to constricted pods (i). Suppose you have crossed YYII parents with yyii parents. Draw the F1 Punnett square and predict the expected F1 phenotype(s).

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Textbook Question

In garden peas, yellow seeds (Y) are dominant to green seeds (y), and inflated pods (I) are dominant to constricted pods (i). Suppose you have crossed YYII parents with yyii parents. List the genotype(s) of gametes produced by F1 individuals.

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Textbook Question

In garden peas, yellow seeds (Y) are dominant to green seeds (y), and inflated pods (I) are dominant to constricted pods (i). Suppose you have crossed YYII parents with yyii parents. Draw the F2 Punnett square. Based on this Punnett square, predict the expected phenotype(s) in the F2 generation and the expected frequency of each phenotype.

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Textbook Question

Imagine repeating the experiment on epigenetic inheritance that is shown in Figure 19.6. You measure the amount of radioactive uridine (U) incorporated into Hnf4a mRNA in counts per minute (cpm) to determine the level of Hnf4a gene transcription in rats born to mothers fed either a normal diet or a low-protein diet. The results are 11,478 cpm for the normal diet and 7368 cpm for the low-protein diet. For this problem, your task is to prepare a graph similar to the one at the bottom of Figure 19.6 that shows the normalized results for the low-protein diet relative to the normal diet. Normalizing values means that the value obtained from one condition is expressed as 1.0 (the norm; the normal diet in this case) and the values obtained from any other conditions (low-protein diet in this case) are expressed as decimal values relative to the norm.

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Textbook Question

The smooth feathers on the back of the neck in pigeons can be reversed by a mutation to produce a 'crested' appearance in which feathers form a distinctive spike at the back of the head. A pigeon breeder examined offspring produced by a single pair of non-crested birds and recorded the following: 22 non-crested and 7 crested. She then made a series of crosses using offspring from the first cross. When she crossed two of the crested birds, all 20 of the offspring were crested. When she crossed a non-crested bird with a crested bird, 7 offspring were non-crested and 6 were crested. For these three crosses, provide genotypes for parents and offspring that are consistent with these results.

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Textbook Question

The smooth feathers on the back of the neck in pigeons can be reversed by a mutation to produce a 'crested' appearance in which feathers form a distinctive spike at the back of the head. A pigeon breeder examined offspring produced by a single pair of non-crested birds and recorded the following: 22 non-crested and 7 crested. She then made a series of crosses using offspring from the first cross. When she crossed two of the crested birds, all 20 of the offspring were crested. When she crossed a non-crested bird with a crested bird, 7 offspring were non-crested and 6 were crested. Which allele is dominant?

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