Saturday, February 26, 2000

(1) A man and a woman have a large family consisting of 10 children, 5 boys and 5 girls. Three of the girls have green noses and three of the boys have green noses. If green noses are inherited as a sex-linked trait, then what are the nose-color phenotypes of the two parents. What are the genotypes of the two parents?

A: The mother has a green nose, the father does not; the green-nose allele is dominant to the wild-type allele; the mother is a heterozygote so approximately half of her offspring received the green-nose allele; We know that the father does not have a green nose because not all of the girl progeny have green noses (which they would have to have if the nose-color locus is X-linked); we know that the nose-color allele isn't Y-linked because the girls are affected as well as the boys (and not all of the boys are affected) (From chapter 15)

(2) About 60% of the mass of ribosomes consists of what non-proteinaceous type of macromolecule? Please be specific.

A: rRNA (From chapter 17)

(3) An organism that possesses three haploid sets of chromosomes is __________. (a description of its karyotype)

A: Polyploid (From chapter 15)

(4) Deaf mutism in human beings is due to the presence of either or both of the completely recessive nonallelic genes a and b in homozygous condition. These genes are located in different autosomes. Give the genotypes of a deaf mute man and a deaf mute woman who can produce only normal children.

A: To answer this question, you must work backward from a normal child; What must a child's genotype be to be normal? A?B?. If both parents are deaf mutes, then each must be recessive at one of these loci, but it must not be the same locus if both can produce only normal children. In fact, one must be homozygous dominant at one locus (which assures that all children will carry at least one dominant allele at that locus) and the other parent must be homozygous dominant at the other locus (which assures that all children will carry at least one dominant allele at the other locus); Thus, the genotypes of the parents must be: AAbb and aaBB which will give rise to offspring that are all: AaBb, i.e., phenotypically normal (From chapter 14)

(5) Determine for each pedigree the method of inheritance of the trait in question and then indicate the genotype of each individual in each pedigree whose genotype you know with certainty.

A: Pedigree "a" is a recessive trait and all affected are homozygous recessive while for the unaffected it is not possible to know any of the genotypes with certainty except the first generation (e.g., the grandparents) for which all must be heterozygous at the locus in question; pedigree "b" is a dominant trait and all affected are heterozygous whereas all unaffected are homozygous recessive (From chapter 14)

(6) Distinguish the definition of the term character from that of the term trait.

A: a character is some feature of an organism whereas a trait is some manner in which a character varies; e.g., eye color is a character and possible traits include brown, blue, green, etc. eyes (From chapter 14)

(7) During what phase of meiosis are Mom’s and Dad’s chromosomes randomly tugged to separate cells?

A: Anaphase I (From chapter 15)

(8) Frameshift mutations can result from which of the following mutation types

            (a) Deletion

            (b) Missense

            (c) Nonsense

            (d) Point

            (e) Silent

A: (a) deletion (From chapter 17)

(9) How many autosomes are found in the human haploid?

A: 22 (46/2 - 1 or 46-2/2… either way it is the number of chromosomes in the diploid divided by two less the number sex chromosomes) (From chapter 13)

(10) Ignoring mutational or chromosomal rearrangement events in somatic cells during development, explain how a woman could be color blind in one eye but have normal vision in the other.

A: X-linked character; woman is a heterozygote; normal-vision X chromosome was turned in the cell lineage that led to one eye while the color blind-vision X chromosome was turned off in the cell lineage that led to the other eye (From chapter 15)

(11) In mice, black color (B) is dominant to white (b). At a different locus, a dominant allele (A) produces a band of yellow just below the tip of each hair in mice with black fur. This gives a frosted appearance known as agouti. Expression of the recessive allele (a) results in a solid color. If mice that are heterozygous at both loci are crossed, what will be the expected phenotypic ratio of the offspring?

A: This is an example of epistasis (and is question number 19 on page 260 of your text); three-quarters of the mice carrying the B allele will be agouti and one-quarter of the mice will be white: 3/4 * 3/4 = 9/16 of the mice will be agouti, 3/4 * 1/4 = 3/16 of the mice will be black, and 1/4 of the mice will be white; the phenotypic ratios will be 9:3:4 (From chapter 14)

(12) In sesame plants, the one-pod condition (P) is dominant to the three-pod condition (p), and normal leaf (L) is dominant to wrinkled leaf (l). Pod type and leaf type are inherited independently. Determine the genotypes for the two parents for a mating that produces the following progeny: 150 one-pod normal, 147 one-pod wrinkled, 51 three-pod normal, and 48 three-pod wrinkled.

A: PpLl x Ppll; note that this problem is perhaps easier than you think; try looking at the different loci independently, i.e., approximately one-half of the progeny are normal and one-half are wrinkled while approximately three-quarters have one pod whereas approximately one-quarter have three pods; the crosses that best explain these results are Ll x ll (which gives rise to one-half dominant phenotype and one-half recessive phenotype) and Pp x Pp (which, of course, is the monohybrid cross); put the two loci back together and you have PpLl x Ppll (this is question number 4d on page 259 of your text) (From chapter 14)

(13) In terms of relative chromosome number and types of nuclear divisions, how does the plant sexual cycle differ from that of humans?

A: Plants display an alternation of generations; that is, their haploid cells are capable of undergoing mitosis (not just no division followed by fertilization as is seen in humans) (From chapter 13)

(14) In terms of their DNA sequences, how (or why?) do the metaphase I (of meiosis) chromatids differ from the interphase I chromatids?

A: The metaphase I chromosomes are a mixture of the chromosomes from mom and dad; thus, the sequence of a given chromosome at interphase I is whatever sequence the chromosome had in dad's sperm or mom's ova; after prophase I, however, the sequence is some combination of the sequence of two homologous chromosomes, the one that came from mom and the one that came from dad (From chapter 13)

(15) In wheat there are three loci for seed color. At each locus there are two alleles, one dominant, one recessive. The presence of any one of these dominant alleles will result in wheat with some color. That is, a white wheat must be homozygous recessive at all three loci. What will be the ratios of F₂ plants with colored kernels to white kernels when the following F₁ plants are allowed to self-pollinate?

            (a) W¹w¹ W²w² W³w³

            (b) W¹W¹ W²w² W³w³

            (c) w¹w¹ W²w² W³w³

A: (a) 0.25 * 0.25 * 0.25 = 1/64 colorless, 63/64 colored; (b) All colored; (c) 0.25 * 0.25 = 1/16 colorless, 15/16 colored (From chapter 14)

(16) Phosphates are attached to what carbon on nucleoside triphosphates?

            (a) 1’

            (b) 2’

            (c) 3’

            (d) 4’

            (e) 5’

            (f) none of the above

A: (e) 5’ carbon (From chapter 16)

(17) Regardless of their dominance relationships at the whole-organismal phenotypic level, two different alleles found in same individual display__________ at the level of molecular gene expression (e.g., as proteins).

            (a) codominance

            (b) complete dominance

            (c) epistasis

            (d) incomplete dominance

            (e) pleiotropy

A: (a) codominance (From chapter 14)

(18) What are X, Y, and Z in the figure?

A: The E site, P site, and A site, respectively (From chapter 17)

(19) Three-part question: define genetic recombination and then name the two processes that contribute to genetic recombination.

A: Genetic recombination is the mixing up of Mom’s and Dad’s genes; the two processes are independent assortment and molecular recombination (From chapter 15)

(20) Two parents are heterozygous for four pairs of alleles on different chromosomes. Each parent is AaBbCcDd. What proportion of their offspring are expected to be AAbbCCDd?

A: 0.25 * 0.25 * 0.25 * 0.5 = 1/128 (From chapter 14)

(21) What are the three substrates of aminoacyl-tRNA-synthases necessary for tRNA activation?

A: tRNA, amino acid, ATP (From chapter 17)

(22) What aspect of DNA replication makes it necessary for linear chromosomes to have telomeres?

A: The requirement for RNA priming makes it impossible to replicate one strand at the end of a linear chromosome (From chapter 16)

(23) What do promoters promote?

A: promoters promote the initiation of transcription (From chapter 17)

(24) What is the frequency of recombination between two loci that are separated by 3 map units?

A: 3% frequency of recombination (From chapter 15)

(25) What is the typical hypothesis proposed as an explanation when a test cross results in less than the expected number of recombinant types (expected meaning as observed by Mendel in his various dihybrid pea crosses)?

A: Linkage (From chapter 15)

(26) On what kind of molecules does one find anticodons? (be specific)

A: anticodons are found on tRNAs (specifically on the anticodon loop); in this case "not specific enough" might be something like "associated with ribosomes during translation" (From chapter 17)

(27) Why must DNA polymerase be capable of stripping RNA off of DNA strands?

A: On the lagging strand DNA polymerization is repeatedly primed by RNA and the completion of Okazaki fragments requires the removal of this RNA and its replacement of with DNA (DNA synthesis is then completed on the lagging strand with the action of DNA ligase) (From chapter 16)