Principles of Biology 1 Fall 2010, HW07

26 Jun Principles of Biology 1 Fall 2010, HW07

Question
Principles of Biology 1

Fall 2010, HW07

1. Caenorhabditis elegans (the famous model genetic organism) has 6 different chromosomes: 5 autosomes (called I-V) and 1 sex chromosome (called X). That is, n=6 for hermaprhodites (which are basically XX females that make both sperm and eggs). Hermaphrodites can either self-fertilize or they can be fertilized by males (which only have one X chromosome per diploid cell; i.e. males are hemizygous and designated “XO”, where the “O” just represents no chromosome, “nullo”).
How many chromatids per nucleus (in a diploid somatic cell) will there be in a hermaphrodite of C. elegans at prophase of mitosis?

A. 6

B. 11

C. 12

D. 22

E. 24

2. How many chromatids per diploid nucleus (in a diploid somatic cell) will there be at telophase of mitosis in C. elegans male cells? (Hint: a cell in telophase contains two separate nuclei.)

A. 6

B. 11

C. 12

D. 22

E. 24

3. During meiosis I of spermatogenesis in C. elegans males, what arrangement will be found at the metaphase plate? (Reminder: a tetrad is a complex of four chromatids; a dyad is a complex of two chromatids.)

A. The autosomes will form tetrads and the X will exist as a dyad.

B. All chromosomes will form tetrads.

C. The autosomes will form tetrads, and the X and O will pair as a tetrad.

D. No tetrads will form in this case because males are hemizygous.

E. The autosomes will form dyads and the X will form a tetrad.

4. After meiosis II of spermatogenesis in males of C. elegans, how many chromatids will each gamete have?

A. Half the gametes will have 5 and half will have 6.

B. All the gametes will have 6 chromatids.

C. All the gametes will have 5 chromatids.

D. One quarter have 5 chromatids, and the rest will have 6 chromatids.

E. Half the gametes will have 11 and half will have 12.

5. During the cell cycle, there are proteins that are involved in degrading cyclins. According to the model in your book (Fig. 12.17), what would happen if such proteins were compromised by mutation such that they could not function well?

A. The concentration of free Cdk would increase, resetting the G2 checkpoint.

B. Less MPF would be made than normal.

C. Phosphorylation of mitosis-inducing molecules would be prevented.

D. Cells would not be “checked” at G2, and the cell cycle could not be stopped.

E. The activity of the cyclin-dependent kinase enzyme would be reduced.

6. In peas, flower color is influenced by gene p. One allele at this locus, P, is dominant and causes flowers to be purplish. Another allele at this locus, p, is recessive and fails to produce purple pigment. You possess a pea plant producing purple flowers, but are unsure of their genotype (except that you know they cannot be homozygous pp). How could you distinguish between the hypotheses that the plant is heterozygous Pp versus homozygous PP?

A. Cross the plant with a known homozygous dominant to see if all the progeny are purple or not.

B. Cross the plant with another purple-flower plant to see if you get any white progeny or not.

C. Cross the plant with a white-flower plant to see if you get any white progeny or not.

D. Cross the plant with a white-flower plant to check for pinkish flowers in the progeny.

E. Dissect the flowers to see if the anthers are all the same or not.

7. Wild-type C. elegans worms are skinny and move in an elegant sinusoidal, curvy pattern, like some snakes. Homozygous dpy(-)/dpy(-) mutants are dumpy (short and fat); homozygous unc(-)/unc(-) mutants are uncoordinated. Assume that the dpy gene segregates independently from the unc gene, and that the wild-type alleles dpy(+) and unc(+) are completely dominant to the mutant alleles dpy(-) and unc(-), respectively. What are the expected results in the F2 generation from selfing an F1 hermaphrodite that came from a cross between a homozygous dumpy-only male (dpy(-)/dpy(-) ; unc(+)/unc(+)) and a homozygous uncoordinated-only hermaphrodite (unc(-)/unc(-) ; dpy(+)/dpy(+))? (Hermaphrodites in this cross behave as females. Hermaphrodite F1s, however, are allowed to self.)

A. 1/16 of the F2 progeny will be both dumpy and uncoordinated.

B. 3/16 of the F2 progeny will be both dumpy and uncoordinated.

C. 9/16 of the F2 progeny will be both dumpy and uncoordinated.

D. All of the F2 progeny will be both dumpy and uncoordinated.

E. None of the F2 progeny will be both dumpy and uncoordinated.

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8. For the pedigree above, assume that the allele responsible for the trait of individuals indicated with colored symbols is T and the wildtype allele is +. What are the genotypes of the two most recently born individuals? (Hint: do not let the symbols fool you.)

A. are T+ and ++.

B. are TT and T+.

C. are TTand ++.

D. are TT for one and either ++ or T+ for the other.

E. cannot be determined for either individual with certainty.

9. First, find out from your book or another resource what is the diploid number of chromosomes for humans. Ignoring recombination, for a normal diploid human couple, what would be the total number of different chromosomal combinations that are possible in zygotes made by the union of their gametes?

A. 23

B. 223

C. 423

D. 446

E. infinte

10. A woman and her spouse both show the normal phenotype for pigmentation, but both had a parent who was albino. Albinism is an autosomal recessive trait. If their first two children have normal pigmentation, what is the probability that their third child will be albino? (No need for sophisticated Bayesian probability calculations; just use normal probability calculations.)

A. 0.00

B. 0.25

C. 0.50

D. 0.75

E. 1.00