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Homework Help: Mendelian Genetic problems

  1. Jun 11, 2008 #1
    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?


    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.
  2. jcsd
  3. Jun 11, 2008 #2
    I figured out the 1st problem, just need help with 2 and 3b.
  4. Jun 11, 2008 #3
    Your part A and B are correct. It looks like you need help with setting up a Punnett Square (PS). We will analyze the eye shape, and you should be able to determine the other on your own. First, let's understand the function of a PS. A PS shows all of the possible combinations of alleles that an offspring could receive from their parents. Now each parent has two alleles for each trait (at least, in the simple cases that you are expected to analyze), and when they reproduce they randomly pass off one of these alleles to their offspring--it's all a game of chance. Therefore, there are only four possible ways that these alleles can combine. Do take note that some of these possibilities will be the same; this is fine, it just means that the offspring has a greater likelihood of having such a trait.
    When setting up a PS, write the parents genotype (AA and aa) as in "PS1.bmp": Their placement does not matter:

    Now, for Parent #1, just write their allele in the same column as it appears. See "PS2.bmp"

    Now, do the same thing for Parent #2, but instead of columns, it's rows. See "PS3.bmp" for the completed PS:

    As is evident form the PS, all four possibilities are the same "Aa", which means that there is a 100% chance that the offspring will show the dominant allele trait, "star eyes", since it masks the recessive allele trait, "square eyes". Now you should be able to do the same for the other trait.

    "Heterozygous" simply means that the individual has one dominant and one recessive allele for a given trait; for example, "Aa" is heterozygous. In contrast, "homozygous" means two of the same alleles; "AA" (homozygous dominant) and "aa" (homozygous recessive) since "A" is the dominant allele and "a" is the recessive allele.
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