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    Introductory Genetics Problem Set

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    1. Two black guinea pigs were mated repeatedly over several years. They produced 29
    black progeny and 9 white progeny. Explain these results, giving the genotypes of
    parents and progeny.

    2. In horses, black is dependent on a dominant gene (B) and chestnut upon its recessive
    allele (b). The trotting gait is due to a dominant gene (T) and the pacing gait due to its
    recessive allele (t). A homozygous black pacer is mated to a homozygous chestnut

    A. What is the genotype of the F1 generation?
    B. What is the phenotype of the F1 generation?
    C. If two F1 individuals were mated, use the fork line method to determine the kinds of
    offspring (phenotypes) would they have and in what proportions?
    D. If an F1 male were mated to a homozygous female black pacer, use the fork line
    method to determine the genotypes and genotypic ratio of offspring that would be

    3. In the cross AaBb X Aabb:
    A. How many different genotypes can occur in the gametes of each parent?
    B. Give the gametic genotypes and gametic ratios for each parent.

    4. Porcupines have a diploid chromosome number (2n) of 34. Considering only
    independent assortment (the random alignment of maternally and paternally derived
    chromosomes), how many genetically different offspring are possible when two
    porcupines mate?

    5. A porpoise has a diploid chromosome number (2n) of 44. How many different
    chromosomal configurations can occur following meiosis?

    6. The amount of DNA in a G1 nucleus of the lemur (Lemur catta) is 6.22 picograms
    (pg). The diploid number of chromosomes in the lemur is 56. What would be the
    number of chromosomes and the DNA content of a nucleus in each of the following:

    Cell Type Number of Chromosomes Amount of DNA (pg)
    Lung cell in G2 ________ ________
    Mitotic metaphase ________ ________
    Second polar body ________ ________
    Sperm ________ ________
    A secondary oocyte ________ ________
    Primary spermatocyte ________ ________

    7. A somatic cell in a garden pea (2n = 14) undergoes mitosis.
    A. How many chromosomes does each progeny cell have?
    B. If a cell in the same plant undergoes meiosis, how many chromosomes does each cell
    have after the first division?
    C. How many chromosomes after the second meiotic division?

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    Solution Preview

    1. When we look at this question, we first look at the ratio of progeny. 29:9 is the actual data, right? This is very similar to a hypothetical 30:10 ratio, isn't it? Therefore, we can say that the theoretical ratio is 3:1. Now, that we know that, we can easily solve the problem.

    In order to get this ratio, each black g.p. must be Bb. Their gametes will be B and b.
    Here's the Punnett Square:

    B b
    B BB Bb
    b Bb bb

    BB and Bb are both black. There are 3 of those. bb is white, there is 1 of those. Therefore, we have a 3:1 phenotypic ratio.


    2. Homozygous black pacer must be BBtt. The homozygous chestnut trotter must be bbTT. We know this because we are told the two horses are homozygous.

    Therefore, we do the crossing. BBtt produces only Bt gametes. bbTT only produces bT as the gametes. Therefore, all the F1 are a combination of these gametes, i.e. BbTt. Simple. What do these horses look like? Since they have the dominant allele in each case, i.e. B and T, the horses are black and trotting.

    In order to figure out the F2 generation, you have to do a 4x4 Punnett Square grid. This is a classic dihybrid cross that is set up as a 4x4 grid. The phenotypic ratios will be 9:3:3:1. You can use these numbers to verify your answer.

    Each BbTt F1 will produce 4 types of gametes: BT, Bt, bT, bt. Make sense? Always make sure you have one of each type of gene, i.e. allele, in gametes.

    You set up the cross like this:

    BT Bt bT bt

    You ...

    Solution Summary

    This solution contains a set of seven introductory genetics problems, may of them having multiple parts. Every question is solved with clear, easy to understand solutions. If you're studying for a test or just beginning to work with genetics, then this solution is the one for you!