1. Consider 3 pairs of homologous chromosomes with centromeres labeled A1/A2, B1/B2, and C1/C2 where the slash line separates one chromosome from its homologue. a). How many different kinds of meiotic products can this individual produce? b). Develop a general formula that expresses the number of different types of gametic chromosomal combinations which can be formed in an organism with k pairs of chromosomes.
   Answer: a). 8; b). 2 raised to the power of k.

2. When a plant of chromosomal type aa pollinates a plant of type AA, what chromosomal type of embryo and endosperm is expected in the resulting seeds?
   Answer: embryo, Aa; endosperm, AAa.

3. At what stage in the cell cycle does chromosome replication occur? At what stage does chromosome replication occur in meiosis?
   Answer: S (synthesis) stage; no chromosome replication during meiosis.
   

4. Three markers are being mapped. The observed recombination frequencies are: d-e, 4.0%; e-f, 2.5%; and d-f, 1.5%. What is the genetic map for these markers?
   Answer: Mapping is based on the additivity of recombination frequencies. The order must be d f e because 1.5 + 2.5 = 4.
   

5. What is the relationship between the amount of DNA in a somatic cell and in a gamete?
   Answer: Twice as much in the somatic cell (2n) than the haploid gamate (n).

6. In what direction do DNA and RNA polymerases move along a template strand?
   Answer: 3' to 5'
   

7. In gel electrophoresis do smaller double-stranded molecules move more slowly or more rapidly than larger molecules? Towards which pole does the DNA migrate (i.e., positive or negative)?
   Answer: Small molecules move faster; they move towards the positive pole because the phosphate backbone of the DNA is negatively charged.
   

8. How do exonucleases and endonucleases differ?
   Answer: Exonucleases cut up the DNA from unprotected ends, whereas the endonucleases cut DNA between ends.

9. Name the bases in DNA. Which of them form base pairs?
   Answer: A, T, G, C: G-C, A-T.

10. What is meant by maternal inheritance?
    Answer: If a trait is carried by a female, in both male x female and reciprocal crosses, the progeny will have the phenotype of the female parent.

11. At one time the possibility was considered that the genetic code might be an overlapping code of the following type: the condons in base sequence CATCATCAT...would be CAT ATC TCA CAT.... How was this hypothesis affected by the observation that mutant proteins usually differ from wildtype protein by a single amino acid?
    Answer: A single base change would usually affect two amino acids.

12. Will a change in the first base of a codon necessarily produce a nonfunctional protein?
    Answer: No, because an amino acid change does not necessarily produce a nonfunctional protein. Also, due to redundancy, the codon may still code for the same amino acid.

13. If you knew the base sequence of a wildtype and a mutant, would you know anything about the dominance or reecessiveness of the mutation?
    Answer: No, because these are phenotypic properties.

14. What term describes a gene that is expressed continually, even though its transcription may be autoregulated?
    Answer: constituitive.

15. What is the advantage of using a plasmid with two antibiotic-resistance genes as a cloning vehicle?
    Answer: One gene could be used as a postive selector for presence of the plasmid, while the second antibiotic-resistance gene could be used as a selector plasmids containing cloned DNA sequences when its knockout is due to the presence of a DNA insert.

16. Give two reasons why a cloned prokaryotic gene might not be expressed in a prokaryote.
    Answer: The gene may have been separated from its promoter or from its ribosome binding site.

17. Summer squash can be found in three shapes: disk spherical, and elongate. In one experiment, two squash plants with disk-shaped fruits were crossed. The first 160 seeds planted from this cross produced plants with fruit shapes as follows: 89 disk, 61 sphere, and 10 elongate. What is the mode of inheritance of fruit shape in summer squash?
    Answer: The numbers are very close to a ratio of 9:6:1, an epistatic variant of the 9:3:3:1, in which the two 3/16ths categories are combined. If this is the case, then the parent plants with disk-shaped fruits were dihybrids (AaBb). Two genes combine to control fruit shape in summer squash and there are epistatic interactins between the two genes giving a (6:1 ratio of offspring phenotypes when dihybrids are crossed.

18. Mendel crossed tall pea plants (TT) with dwarf ones (tt). The F1 plants were all tall. When these F1 plants were selfed to produce the F2 generation, he got a 3:1 tall to dwarf ratio of offspring. Give the genotypes and phenotypes and relative proportions of the F3 generation produced when the F2 generation was selfed.
    Answer: Dwarf F2 (1/4 of total F2), when selfed, produce all dwarf progeny (tt). Tall F2 (3/4 of total F2), when selfed, fall into two categories: 1/3 (TT, 1/4 of total F2) produces all tall, and 2/3 (Tt, 1/2 of total F2) produces tall and dwarf progeny in a 3:1 ration. (The 3:1 ratio is from 1/2 the F2, so the tall component is 3/8 of the total F3 [3/4 * 1/2] and the dwarf is 1/8 of the total F3 [1/4 * 1/2]). Overall, the F3 are 3/8 TT (tall), 2/8 Tt (tall), and 3/8 tt dwarf.

19. In one sentence or two, state Mendel's two "laws."
    Answer: The two alleles of a gene separate into different gametes, and this assortment is independent of the assortment of alleles of other genes.

20. What are Southerns, Northerns and Westerns?
    Answer: Southerns are a DNA hybridization method in which, following electrophoretic separation, denatured DNA is transferred from a gel to a paper filter and then exposed to labeled DNA or RNA probes under conditions of renaturation.
    Northerns are a nucleic acid hybridization method in which RNA is electrophoretically separated by size, blotted to a filter and probed.
    Westerns are a protein hybridization method in which protein is electrophoretically separated by size and/or charge, blotted and probed with a labeled antibody.
    Non-radioisotopic labeled probes for Southerns and Northerns use antibody recognition of the labeling ligand.

21. What is transposon-tagging?
    Answer: The identification of a DNA sequence by the physical proximity (insertion of linkage) of a transposable to that DNA sequence.

22. What is insertional mutagenesis?
    Answer: The inactivation or disruption of normal function of a gene by the insertion of a transposable element or foreign DNA into a critical region of a gene, such as an exon, exon-intro border, or regulatory region.

23. What does RFLP" an acronym for?
    Answer: restriction fragment length polymorphism.
    

24. Name at least five factors lead to deviations from Mendel's laws.
    Answer: linkage, expressivity, penetrance, epistatis, lethality, incomplete dominance

25. In pea plants, tall growth habit (Le) is dominant to dwarf (le), tendrils (Tl) is dominant to tendrilless (tl), and round cotyledons (R) is dominant to wrinkled (r). A homozygous wildtype inbreed plant is crossed with a dwarf, tendrilless, wrinkled-seed variety. What types of gametes will be produced by the resulting F1 plants, and in what relative proportions a) assuming independent segregation b) assuming tight linkage between Tl and R?
    Answer: The homozygous wildtype parent has the genotype of Le/Le Tl/Tl R/R. The other parent has the genotype of le/le tl/tl r/r. All gametes produced by the wildtype parent are Le Tl R, and all gametes produced by the other parent are le tl r -- regardless of the linkage relationship. Therefore, the genotype of the F1 individuals are Le/le Tl/tl R/r.
    a) Assuming independent segregation, the F1 plants will produce (2)(2)(2) = 8 types of gametes in equal proportions:
    
    • Tl Le R
      
    • Tl Le r
      
    • Tl le R
      
    • Tl le r
      
    • tl Le R
      
    • tl Le r
      
    • tl le R
      
    • tl le r
      
    b) Assuming linkage between Tl and R, the F1 plants will produce (2)(2) =4 types of gametes in equal proportions:
    
    • Tl Le-R
      
    • tl Le-R
      
    • Tl le-r
      
    • tl le-r

26. A true-breeding wildtype Arabidopsis thaliana plant (Ap1/Ap1)was crossed to the flowering mutant apetala1(ap1/ap1). Sixteen F2 seed were planted, and the first twelve were wildtypes (Ap1/-). What is the probability that the remaining four plants will be the mutant (ap1/ap1)?
    Answer: (1/4) (1/4) (1/4) (1/4) = 1/256.

27. Name two differences between RNA and DNA.
    Answer: RNA is single-stranded and has a Uracil in place of Thymydine, whereas DNA is double-stranded and is composed of A, G, C and T.
    

Problems 3-16 from Hartl, DL, Freifelder, D and LA Snyder. (1988) Basic Genetics. Jones and Bartlett, Publishers. pp 505.

Problems 17 and 18 from Tamarin, RH. (1996) Principles of Genetics, fifth edition. Wm. C. Brown Publishers, E. M Sievers, Ed. Pp 683.

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