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genetic homework help. Thanks

Genetics as it applies to evolution, molecular biology, and medical aspects.

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genetic homework help. Thanks

Postby undergradmember » Mon Feb 28, 2011 12:16 am

due to the role of Drosophila as a biological model, many eye color mutants have been isolated and used to analyze gentic control of eye pigmentation. Wild-type eye color is brick red due to a mixture of red and brown pigment. Vermillion (X^v) is a recessive, X linked mutation causing bright red eye color due to lack of brown pigments. Brown (b) is a recessive, autosomal muation causing brown yee color due to lack of red pigments. white eyes result from the loss of both red and brown pigments. crosses between vermillion and brown flies show that the two mutation complement each other.

A) what genotype(s) and eye color phenotype(s) do you expect to see in progeny of a cross between a brown female and vermillion male (both from true breeding lines)? if male and female progeny would look different, clearly indicate this.

B) in a completely sepated cross, half progeny are brown eyed and half the progeny are white eyed. what are the genotype and phenotype of the parents?

my answers:

Part A)

according to the given informations:

wildtype = brick red eyes = red + brown

vermillion = red and lacks off brown; therefore to be vermilion individual must be either X^v X^v Bb or X^v Y B_

Brown = lacks of red and bb; therefore X^+ X^+ bb or X^+Y bb

white= lacks of both red and brown so X^+ X^v B_ or X^+ Y B_



we have:

female brown crosses with vermilion male

genotypes for them are:

female male
XX^+ X^+ bb X X^v Y B_(B or b) but it says that these guys are true breeding so! it is capital B

the cross will yield

female: X^v X^+ Bb red brick

male: X^+ Y Bb red brick also


Part B)

working backward from offspring to find out parents genotypes and phenotypes

we know half progeny are brown eyed and half the progeny are white eyed.

therefore

X^+ X^+ bb x X^+ Y Bb or X^+ X^+ Bb x X^+ Y bb i dont think it change anything. always, this lead to the parents must be:

X^+ X^+ bb (brown female) and X^+ Y Bb (white male) or X^+ X^+ Bb(white female) x X^+ Y bb (brown male)


please tell me if i did it correctly



NEXT QUESTION>

2)In both Manx and Maine Coon cats, animals may be bushy tailed or tail less. The Manx animals have shorter hind legs than front legs whereas the Maine Coon animals have same length legs. When many offspring were obtained from crosses between bushy tailed Maine Coons and tail less Manx cats, half were found to be bushy tailed Manx and half were tail less Maine Coons. when these tow types of F1 cats were mated to one another, the following F2 data were collected:

3/8 Bushy Tailed Manx 3/8 Bushy Tailed Maine Coons
1/8 Tail less Manx 1/8 Tail less Maine Coons

a) A geneticist knew that Maine Coons are true breeding while Manx were not. Maine Coons produce more offspring than Manx. Provide genetic explaination for these observation.

I have no idea was is going on but I think it is due to Lethal Alle Recessive trait. please tell me if i am wrong and if i miss something else.

b) Manx kittens are occasionally born with partial tails called stumpy and half tails called stubby. give a possible explanation for this.

Because Manx cats were not true breeding, i believe it might be the result of co- dominance?
please let me know.



Thank you for reading my long hw problems. I really appreciate you guys for taking your time to read this.
Thank you again.
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Postby DRT23 » Wed Mar 02, 2011 2:56 pm

Hi undergradmember,

I'll write my answer for question 1 here. But, I can't say whether my answer or your answer is correct since I'm also an undergrad and can not guarantee my answer :)

> I think 'complement mutation' means that the mutations are related to the same trait. So, this info is not so important for answering this question. It's a kind of extra information. (If I'm wrong, please let me know).

> 'True breeding' means the organism is homozygous for the trait(s).

I think:

Vermillion=
X^v X^v + BB
X^v X^v + Bb
X^v Y + BB
X^v Y + Bb

Brown=
X^V X^V + bb
X^V X^v + bb
X^V Y + bb

(X^V is other, the dominant, allele for vermillion's loci and is wildtype)

White=
X^v X^v + bb
X^v Y + bb

Therefore, the answer for A:

A Brown Female=
X^V X^V + bb
X^V X^v + bb

A Vermillion Male=
X^v Y + BB
X^v Y + Bb

(See details for each combination above) As they must be true breedings, we narrow posibilities (only homozygotes):

A Brown Female=
X^V X^V + bb

A Vermillion Male=
X^v Y + BB

So, gamets from mother carry X^V + b only and gamets from father carry X^v + B or Y + B. Thus, the F1 is:

X^V X^v + Bb (all females are wildtype for eye)
X^V Y + Bb (all males are also wildtype for eye)
And male and female progeny would look the same (wildtype).

Answer for B:

X^V X^v + bb with X^v Y + bb (brown female with white male)
or
X^v X^v + bb with X^V Y + bb (white female with brown male)

After writing my answer and looking back at your answer, I saw the two answers are the same :) So, you did it correctly.
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Postby DRT23 » Wed Mar 02, 2011 4:01 pm

I've been thinking about your second question for a while. But, this one is more difficult and I can't find any answer. The reason for Manx cats' producing less offspring may be lethal allele recesive. But, I can't find genotypes for these cats because it says Maine cats are true breeding, but Manx aren't. But, it contradicts the given data. If 3 of 4 Manx are bushy tail and 1 of 4 is tailles, I think 1 AA and 2 Aa are bushy and aa is tailles. But if Manx cats are not true breeding, there will be only Aa heterozygouts (bushy tail) and no tailles Manx cat. Am I wrong?

As you see, I'm confused with the question. So, what do you think? Did you learn the answer? If you did or when you did, please share it with us.
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Re:

Postby undergradmember » Wed Mar 02, 2011 11:26 pm

DRT23 wrote:I've been thinking about your second question for a while. But, this one is more difficult and I can't find any answer. The reason for Manx cats' producing less offspring may be lethal allele recesive. But, I can't find genotypes for these cats because it says Maine cats are true breeding, but Manx aren't. But, it contradicts the given data. If 3 of 4 Manx are bushy tail and 1 of 4 is tailles, I think 1 AA and 2 Aa are bushy and aa is tailles. But if Manx cats are not true breeding, there will be only Aa heterozygouts (bushy tail) and no tailles Manx cat. Am I wrong?

As you see, I'm confused with the question. So, what do you think? Did you learn the answer? If you did or when you did, please share it with us.



i havent find out the answer yet. but i am going to maybe later this week like Friday! I hope i did it right or 20 points off!!! Thanks for reading my thread.
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Postby DRT23 » Thu Mar 03, 2011 4:43 pm

You're welcome! It was a pleasure for me to think about these questions. And please post the answer when you get it. I really wonder it.
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