Mendelian Genetic problems

[Ph0bolus]

My instructor gave me some problems to help study for the quiz on Monday, but i have no idea how to go about solving them..

 

1) In martians, Gene 1 is determining eye shape (A-star eyes; a- square eyes) and Gene 2 is coding for presence or absence of tail hairs (B-hairy tail; b-smooth tail). True- breed star eyed individual with smooth tail marries another true-breed square-eyed individual with hairy tail and all of their childred have star eyes and hairy tailes. If mendelian Genetics works on Mars,

 

a) which allele of each gene is dominiant, and which is recessive?

 

Would the dominant ones be A,B, and the recessive a,b

 

b) What are the genotype of the parents?

 

Would the mother be A/A b/b and the father a/a B/B

 

c) What combinations of genes and in what proportions do you expect in the resulting gametes in parents?

 

I don't know how to figure this one out.

 

2)Assume that D,E,F,G,H and I are genes on ifferent chromosomes. From the mating (parent A) DdeeFfGGHhIi * (parent B) DdEEFFGgHhii:

 

a) What is probability that one of the offspring will have the genotype DdEeFFGghhIi?

 

would this answer be 1/(2^7)?

 

b) What is the probability that one of the offspring will be heterozygous for each allele?

 

I'm lost on this one.

 

c) How many different kinds of gametes can be produced by each parent?

 

128?

 

3)A dominant allele of a gene "A" causes yellow color in rats. The dominant allele of another independent gene, "R", produces black coat color. When 2 dominant genes occur together (A-R-), they produce grey coat color. Rats of the double recessive genotype are cream colored.

 

a) What kind of gene interaction iss that?

 

Would this be incomplete dominance?

 

b)If a grey male and a yellow female mated and produced approx. 3/8 yellow, 3/8 grey, 1/8 cream, and 1/8 black, what were the genotypes of the parents?

 

I tried the punnet squares, but keep getting confused with it.

 

Any help would be greatly appreciated, Thanks in advance.

[CharonY]

Well, start off with the genotype of the parents. You know their phenotypes, so you just have to associate each phenotype (eyes and tail type) with the given genotype (aA or Bb).

[Dak]

1) In martians, Gene 1 is determining eye shape (A-star eyes; a- square eyes) and Gene 2 is coding for presence or absence of tail hairs (B-hairy tail; b-smooth tail). True- breed star eyed individual with smooth tail marries another true-breed square-eyed individual with hairy tail and all of their childred have star eyes and hairy tailes. If mendelian Genetics works on Mars,

 

a) which allele of each gene is dominiant, and which is recessive?

 

Would the dominant ones be A,B, and the recessive a,b

 

yuppers.

 

made so much difficulter by the fact they used standard notation for dominance/recessivity (Aa, Bb)

 

 

b) What are the genotype of the parents?

 

Would the mother be A/A b/b and the father a/a B/B

 

 

why not the other way round?

 

c) What combinations of genes and in what proportions do you expect in the resulting gametes in parents?

 

I don't know how to figure this one out.

 

actually, i'm not sure what exactly this means tbh...

 

2)Assume that D,E,F,G,H and I are genes on ifferent chromosomes. From the mating (parent A) DdeeFfGGHhIi * (parent B) DdEEFFGgHhii:

 

a) What is probability that one of the offspring will have the genotype DdEeFFGghhIi?

 

would this answer be 1/(2^7)?

 

yup

 

b) What is the probability that one of the offspring will be heterozygous for each allele?

 

I'm lost on this one.

 

well, for the first gene, the parents are Dd * Dd; that gives, in equal likelyhood, DD, Dd, dD, dd. two of them are hetrozygous, so the chance is 2 * (1/4) = 0.5 for the first gene.

 

c) How many different kinds of gametes can be produced by each parent?

 

128?

 

no, miles off.

 

a gamete will have one allele for each gene. so, a parent that is Aa can make 'A' and 'a' gametes.

 

a parent that's AaBbCC can make ABC, AbC, aBC, abC = 4 different types of gamete.

 

3)A dominant allele of a gene "A" causes yellow color in rats. The dominant allele of another independent gene, "R", produces black coat color. When 2 dominant genes occur together (A-R-), they produce grey coat color. Rats of the double recessive genotype are cream colored.

 

a) What kind of gene interaction iss that?

 

Would this be incomplete dominance?

 

no. incomplete dominance is across one gene. so, if R = red and w = white, then:

 

RR will be enough red pigment to make the phenotype red

ww will be no pigment, == white

Rw will either make enough pigment to make the phenotype red (dominance), or enough to make it pink (incomplete dominance). if it was on one gene and, say, RR == red, YY == yellow, but RY == orange, then it'd be co-dominance.

 

But, this involves more than one gene...

 

b)If a grey male and a yellow female mated and produced approx. 3/8 yellow, 3/8 grey, 1/8 cream, and 1/8 black, what were the genotypes of the parents?

 

I tried the punnet squares, but keep getting confused with it.

 

 

give up on this one till you know the answre to the last question.

[DeanK2]

The later questions require a slight knowledge in the dihybrid principle (Mendel) and co -dominance. If you need an explanation of both, with punnet squares, simply post.