Mendelian Genetic problems

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Mendelian Genetic problems

Post by Alla on Fri Jun 13, 2008 3:34 am

Ph0BoLuS wrote: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.


First of all, you answered 1a and 1b correctly.

1c) What combinations of genes and in what proportions do you expect in the resulting gametes in parents?

You need to write out genotypes of the parents:

Parent 1: AAbb and Parent 2: aaBB (since they are true breeding they are homozygous)

Now you can easily see that each parent can have 1 gamete combination: Ab for parent 1 and aB for parent 2.

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

a) What is probability that one of the offspring will have the genotype DdEeFFGghhIi?
First you need to calculate probabilities for each gene pair separately (if you need to understand where the numbers are coming from you need to make a punnett square for each gene: DdxDd produce DD, Dd, and dd. 2/4=0.5 probability of Dd genotype etc.):

Dd 0.5; Ee 1.0; FF 0.5; Gg 0.5; hh 0.25; Ii 0.5

To get the probability of having this combination you need to multiply all of individual probabilities:
0.5x1.0x0.5x0.5x0.25x0.5=

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

Same thing as above. You do individual probabilities for heterozygous:
Dd 0.5; Ee 1.0; Ff 0.5; Gg 0.5; Hh 0.05; Ii 0.5 and then multiply them.

c) How many different kinds of gametes can be produced by each parent?
(parent A) DdeeFfGGHhIi homozygous for e and G, so only 4 genes are heterozygous.
(parent B) DdEEFFGgHhii homozygous for E, F, and i, so only 3 genes are heterozygous in this case.

To get the number of gametes you need to calculate 2^n where n is the number of heterozygous genes:
Parent A: 2^4=16; Parent B is

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 is that?

I would say epistasis gene gene interaction. Incomplete dominance refers to alleles of the same gene, not 2 different genes as in this case.

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?

No real need for punnett squares here.

Gray male is A_R_ and Yellow female is A _ rr

Now, to get yellow offspring both parents need to have at least one r allele, so that means that Gray male has to be A_Rr.

To get cream offspring that is double recessive aarr you need to need both parents to have at least one a allele, so Gray male is AaRr and Yellow female is Aarr. If you do a punnett square now using these genotypes you will get the given ratios.


Good luck!

P.S. I would appreciate it if you register and use this website directly for your future questions.

Alla
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