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Biology 2112: Lab 11 Genetics Problems Fall 2015 These problems are the final lab exercise and are worth 15 points. Similar problems will appear on...

                        RrGg               rrGG                RrGG              rrgg     

2.  What progeny would you expect in the following crosses?  Choose and recombine the gametes. You do not need to have a complete Punnett Square showing repeated possibilities. 

            RrGg x RrGG             Rrgg x rrGG                RrGG x Rrgg              RrGg x rrgg                

3.  In Tomatoes, the allele for red fruit is dominant over the allele for yellow fruit.  If a plant that is homozygous for red fruit is crossed with a plant with yellow fruit –             

     a) what is/are the genotype(s) and the phenotype(s) of the fruit produced in the F1 generation?

     b) what is/are the genotype(s) and the phenotype(s) of the fruit produced in the F2 generation?

     c) what is the phenotypic ratio of the F2  fruit?

4.  A plant with red flowers is crossed with a plant with white flowers.  All the F1 plants produced red flowers.  When the F1 plants are allowed to self-pollinate, they produce 623 plants with red flowers and 198 plants with white flowers.  What are the genotypes of the P and F1 generations?   

5.  In Tomatoes, the allele for red fruit is dominant over the allele for yellow fruit.  The allele for tall plants is dominant over the allele for short plants.  Find the genotypes and phenotypes of the F1 and F2 generations and the phenotypic ratio of a cross between a homozygous dominant in both genes and a homozygous recessive in both genes. The homozygous individuals are the Parental generation.

6. The probability that a baby will be a boy is ½ as is the probability that a baby will be a girl.  Explain this fact by explaining the mechanism of meiosis in the production of gametes and the process of fertilization.  If a family has 4 boys and 3 girls, what is the probability that the next child will be a girl?

7. Duchenne Muscular Dystrophy is caused by a relatively rare X-linked recessive allele (mutation). Use XM for a normal gene and Xm for the mutated allele. It causes progressive muscle degeneration and usually leads to death by age 20.  An unaffected man marries a normal woman whose brother was affected by this disorder.  Neither set of parents of the couple have muscular dystrophy. What is the probability their first child will be a son who is affected with the disease, how about a daughter?  Explain, using a Punnett square(s), how you arrived at your answer.          

8.  A skin cell from a human female undergoes mitosis. For an unknown reason, the centromeres of one of the two number 4 chromosomes fail to separate. Describe the resultant chromosome number of the two daughter cells.  Indicate the genetic make-up each cell.  Suppose this happened to a single cell in reproductive tissue undergoing meiosis II (the cell went through Meiosis I normally). What would be the chromosome number of the resulting cells from this cell?  Provide a diagram to illustrate this.  If instead of a problem with Meiosis II, the problem was with homologous chromosomes not separating in Meiosis I, what would be the complement of chromosomes for each of the resultant gametes produced?  Provide a diagram to illustrate this.  For all scenarios assume that the cells finish mitosis and meiosis even if not viable due to an abnormal chromosome count.   

9.   A rare condition known as adermatoglyphia leads to such smooth fingertips that the individual has no fingerprints. It has been dubbed the "immigration delay disease" because sufferers have such a hard time entering foreign countries. Recently the cause has been traced to a point mutation in the very first nucleotide of an intron.  The allele with this mutation (F) is dominant to the wild type allele (f).   The condition also leads to less hand sweat than the average person and researchers think that the gene might help skin cells fold over one another early in fetal development.   A woman with adermatoglyphia marries a man with adermatoglyphia (assume both are heterozygous).  Why might it be difficult to calculate the probability they will have a child with fingerprints. Consider the genotypes and phenotypes of all potential children and the possible contribution of the gene to early fetal development?  (http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3155166/ -  Amer. J. Hum. Genet. 89, 302–307, 2011. You do not need to consult this reference but you should read a commentary on it http://news.sciencemag.org/sciencenow/2011/08/the-mystery-of-the-missing-fingerprints.html).

    Note that this problem is as much to teach you about the ambiguities of some problems as statistics.

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10.  As a consequence of independent assortment of chromosomes during meiosis (the biological explanation for Mendel’s 2nd law), each mature gamete from an individual differs greatly in the combination of chromosomes inherited from the parents of that individual.  However, random assortment of chromosomes is not the sole source of reassorting genetic information during meiosis.  Explain why this could be so using the following data.  On the same chromosome there are two genes for finger length and hair texture, with long fingers (F) & curly hair dominant (H) over short fingers (f) and straight hair (h).  The parents of a man had the following genotypes FFHH and ffhh.  His wife shows the homozygous recessive phenotype for both traits, short fingers with straight hair.  One of their children has the long finger and straight hair phenotype.  Diagram the homologous chromosomes, the gametes, any relevant events before and during gamete production and fertilization to support your argument explaining how this child could have this particular phenotype.

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