genetics exam 2 ch 7 and 5 hw questions

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genetics exam 2 ch 7 and 5 hw questions
2014-11-10 04:14:05

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  1. C2. Conjugation is sometimes called
    “bacterial mating.” Is it a form of sexual
    reproduction? Explain.
    It is not a form of sexual reproduction, in which two distinct parents produce gametes that unite to form a new individual. However, conjugation is similar to sexual reproduction in the sense that the genetic material from two cells are somewhat mixed. In conjugation, there is not the mixing of two genomes, one from each gamete. Instead, there is a transfer of genetic material from one cell to another. This transfer can alter the combination of genetic traits in the recipient cell.
  2. C4. What is the difference between an F + and an Hfr strain? Which type of strain do you expect to transfer many bacterial genes to
    recipient cells?
    An F+ strain contains a separate, circular piece of DNA that has its own origin of transfer. An Hfr strain has its origin of transfer integrated into the bacterial chromosome. An F+ strain can transfer only the DNA contained on the F factor. If given enough time, an Hfr strain can actually transfer the entire bacterial chromosome to the recipient cell.
  3. What is the role of sex pili during conjugation?
    Sex pili promote the binding of donor and recipient cells, and provide a passageway for the transfer of genetic material from the donor to the recipient cell.
  4. C10. (185) What is co-transduction? What determines the likelihood that two genes will be co-transduced?
    Cotransduction is the transduction of two or more genes. The distance between the genes determines the frequency of cotransduction. When two genes are close together, the cotransduction frequency would be higher compared to two genes that are relatively farther apart.
  5. C16. A.What form of genetic transfer (i.e., conjugation, transduction, or transformation) would be the most likely mechanism of
    interspecies gene transfer? 

    B. Because S. aureofaciens is a nonpathogenic soil bacterium and E. coli is an enteric bacterium, do you think it was direct gene transfer, or do you think it may have occurred in multiple steps (i.e., from S. aureofaciens to other bacterial species and then to E. coli)?

    C. How could the widespread use of antibiotics to treat diseases have contributed to the proliferation of many bacterial species that are resistant to antibiotics?
    • A. If it occurred in a single step, transformation is the most likely mechanism because conjugation does not usually occur between different species, particularly distantly related species, and different species are not usually infected by the same bacteriophages.
    • B. It could occur in a single step, but it may be more likely to have involved multiple steps.
    • C. The use of antibiotics selects for the survival of bacteria that have resistance genes. If a population of bacteria is exposed to an antibiotic, those carrying resistance genes will survive and their relative numbers will increase in subsequent generations.
  6. E4. What is an interrupted mating experiment? What type of experimental information can be obtained from this type of study? Why is it necessary to interrupt mating?
    An interrupted mating experiment is a procedure in which two bacterial strains areallowed to mate, and then the mating is interrupted at various time points. The interruptionoccurs by agitation of the solution in which the bacteria are found. This type of study isused to map the locations of genes. It is necessary to interrupt mating so that you can vary the time and obtain information about the order of transfer; which gene transferred first,second, and so on.
  7. An Hfr strain that is hisE+ and pheA+ was mated to a strain that is hisE- and pheA-. The mating was interrupted and the percentage of recombinants for each gene was determined by streaking onmedia that lacked either histidine or phenylalanine. The following results were obtained:

    A. Determine the map distance (in minutes) between these two genes. 

    B. In a previous experiment, it was found that hisE is 4 minutes away from the gene pabB. PheA was shown to be 17 minutes from this gene. Draw a genetic map describing the locationsof all three genes.
    • A. If we extrapolate these lines back to the x-axis, the hisE intersects at about 3 minutesand the pheA intersects at about 24 minutes. These are the values for the times of entry. Therefore, the distance between these two genes is 21 minutes (i.e., 24 minus 3).
    • B. --hisE--4--PabB-------17--------pheA
  8. E14. In an experiment involving P1 transduction, the cotransduction frequency was 0.53. How far apart are the two genes?
    • Cotransduction frequency = (1 – d/L)3
    • 0.53=(1/2 minutes)3
    • (1/2 minutes)=cuberoot(.53)
    • (1/2 minutes)=0.81
    • d=.38 min
  9. C2. Describe the inheritance pattern of maternal effect genes. Explain how the maternal effect occurs at the cellular level. What are the expected functional roles of the proteins that are encoded by maternal effect genes?
    • In maternal effect, the mother’s genotype determines the
    • phenotype of the offspring. In the cellular level, the oocytes in the mother
    • are maturing and they are being given nutrients by nursing cells. During this
    • process the nursing cells gives the gene product which persists as the embryo
    • is developing. The proteins are expected to function as structural proteins
    • needed during development of the embyo.
  10. C6. Suppose a maternal effect gene exists as a normal dominant allele and an abnormal recessive allele. A mother who is phenotypically abnormal produces all normal offspring. Explain the genotype of the mother.
    The genotype of the mother would be heterozygous. Since the offspring is normal, this means the mother’s father gave the Dominant allele and the mother’s mother have homozygous recessive and maternal effect influence the genotype.
  11. C10. With regard to the numbers of sex chromosomes, explain why dosage compensation is necessary.
    Dosage compensation is necessary females have two of the same sets of chromosomes. In order to have only one of them express the genes.
  12. C14. Describe the molecular process of X inactivation. This description should include the three phases of inactivation and the role of the Xic. Explain what happens to X chromosomes during embryogenesis, in adult
    somatic cells, and during oogenesis.
    • The first step is targeting the X chromosomes for inactivation. The second step is the transcription of the Tsix gene which
    • blocks transcription of the Xist gene of the targeted X chromosome. The last step is condensing of the silenced gene into a barr body. In somatic cells the same X-inactivation is maintain and during oogenesis, the barr body is uncondensed in order to transmit either copy of the X chromosome.
  13. C16. How many Barr bodies would you expect to find in humans with the following abnormal compositions of sex chromosomes?
    A. XXY 
    B. XYY
    C. XXX
    D. X0 (a person with just a single X chromosome)
    • A. one 
    • B. zero
    • C. Two
    • D. Zero
  14. C18. A black female cat (XB XB)
    and an orange male cat (Xo Y)
    were mated to each other and produced a male cat that was calico. Which sex
    chromosomes did this male offspring inherit from its mother and father?
    Remember that the presence of the Y chromosome determines maleness in mammals.
  15. C24. How is the process of X inactivation similar to genomic imprinting? How is it different?
    • They are both similar in that they both occur during embryonic development. They are different in X-inactivation in mammals is not
    • sex dependent. In X-activation, no marking process occurs during developing. This is not the case in genomic imprinting.
  16. E6. On a recent camping trip, you find one male snail on a deserted island that coils to the right. However, in this same area, you find several shells (not containing living snails) that coil to the left. Therefore,
    you conclude that you are not certain of the genotype of this male snail. On a different island, you find a large colony of snails of the same species. All of these snails coil to the right, and every snail shell that you find on this second island coils to the right. With regard to the maternal effect gene that determines coiling pattern, how would you determine the genotype of the male snail that you found on the deserted island? In your answer, describe your expected results.
    • I would mate the snail the nails in the deserted with a
    • snail in the other island and assign them as F1. The offspring that are made
    • are assigned as F2 and then have them mate with each other. By assuming that
    • the snail from that large colony is homozygous dominant, any offspring of F2
    • (F3) that will have a shell coiled to the write will indicate that the snail in
    • the deserted island is heterozygous.
  17. E8.  Two male mice, which we will call male A and male B, are both phenotypically normal. Male A was from a litter that contained half
    phenotypically normal mice and half dwarf mice. The mother of male A was known to be homozygous for the normal Igf 2 allele.
    Male B was from a litter of eight mice that were all phenotypically normal. The parents of male B were a phenotypically normal male
    and a dwarf female. Male A and male B were put into a cage with two female mice that we will call female A and female B. Female A
    is dwarf, and female B is phenotypically normal. The parents of these two females were unknown, although it was known that they were from the same litter. The mice were allowed to mate with each other, and the following data were obtained: Female
    A gave birth to three dwarf babies and four normal babies. Female B gave birth to four normal babies and two dwarf babies. Which
    male(s) mated with female A and female B? Explain.
    Male B mated with both females because Male A is homozygous for normal lgf2. This is because the mother of Male A is also homozygous for normal lgf2 and the father must have been heterozygous because half of the litter was dwarf. Male B can be lgf2+ lgf2- or lgf2+ lgf2+ because father male be was lgf2+ lgf2+ since all of the litter had a normal phenotype. Since the mother of male B was a dwarf, we know there was a chance that there was a chance that it could have passed an lgf2- allele making male B lgf2+ lgf-.