Genetic exam III

At the molecular level, what are alleles?

Alternative DNA sequences

Which of the following is NOT true of chromosomal duplications? - They are usually lethal when homozygous. - They can be caused by unequal crossing-over. - They play an important role in the evolution of paralogs. - They can be caused by X-ray radiation. - They form visible loops during mitosis and meiosis.

- They are usually lethal when homozygous.

What would the (A+T)/(C+G) ratio be if the tetranucleotide hypothesis was correct?

1

Enzymatic digestion of chromatin DNA using DNase results in how many nucleosomes in the 600 bp fragment?

3

In the famous Hershey-Chase experiment, bacteriophage (phage for short) containing either radioactive sulfur (35S) or phosphorus (32P) were used to determine whether DNA or protein was the genetic material of phage. What was the result of the Hershey-Chase experiment?

32P was detected in infected E. coli cells and the phage that emerge from infected cells, suggesting that DNA was the genetic material.

the results of DNase digestion of chromatin are shown. How many nucleosomes are present in the ~800 base-pair (bp) band would you expect?

4

Why is a random mutation more likely to be deleterious than beneficial?

A gene is likely to be the product of perhaps a billion or so years of evolution. Each gene and its product function in an environment that has also evolved, or coevolved. A coordinated output of each gene product is required for life. Deviations from the norm, caused by mutation, are likely to be disruptive because of the complex and interactive environment in which each gene product must function. However, on occasion a beneficial variation occurs.

Meselson and Stahl used the heavy nitrogen isotope 15N to test the prevailing models of DNA replication. They allowed bacteria with 15N-labeled DNA to replicate their DNA twice in medium containing exclusively 14N ("light") nitrogen sources. They then used equilibrium density gradient centrifugation on DNA extracted after both the first and second replication cycles. (Reminder: heavy 15N = high density, light 14N = low density). If DNA replication was dispersive, what would you expect for the second replication cycle?

One band of intermediate-to-low density.

If DNA replication was dispersive, what banding pattern would you expect in Generation 2 from the Meselson-Stahl experiment?

One band of pure 15N/15N

What might Watson and Crick have concluded had Chargaff's data from a single source indicated the following? (percent of base pairs) A: 29% T: 19% G: 21% C: 31% Why would this conclusion be contradictory to Wilkins's and Franklin's data?

Because in double-stranded DNA, A-T and G-C (within limits of experimental error), the data presented would have indicated a lack of pairing of these bases in favor of a single- stranded structure or some other non-hydrogen-bonded structure. Alternatively, from the data it would appear that A = C and T = G, which would negate the chance for typical hydrogen bonding since opposite charge relationships do not exist. Therefore, it is quite unlikely that a tight helical structure would form at all. In conclusion, Watson and Crick might have concluded that hydrogen bonding is not a significant factor in maintaining a double-stranded structure.

You have identified a novel antibiotic and want to test it for mutagenicity using the Ames Test. You obtain the following results: (Sample : Number of His+ revertant colonies) saline: 7 rat liver enzymes + saline: 9 antibiotic + saline: 103 antibiotic + rat liver enzymes in saline: 212 What conclusion is most consistent with this data?

Both the antibiotic and its metabolites are mutagenic.

Which model of replication did the first generation disprove from the Meselson-Stahl experiment?

Conservative

Phoebus Levene's tetranucleotide hypothesis, if true, would have eliminated the possibility that DNA was the genetic material because...

DNA would not possess the nucleotide sequence diversity required to store information

Describe the sequence of research findings that led to the development of the model of chromatin structure.

Digestion of chromatin with endonucleases, such as micrococcal nuclease, gives DNA fragments of approximately 200 base pairs or multiples of such segments. Digestion for longer times revealed shortened DNA fragments (147 bp) and suggested the presence of linker DNA. Regularly spaced beadlike structures (nucleosomes) were identified by electron microscopy. X-ray diffraction data indicated a regular spacing of DNA in chromatin.

Why is DNA synthesis expected to be more complex in eukaryotes than in bacteria? How is DNA synthesis similar in the two types of organisms?

Eukaryotic DNA is replicated in a manner that is very similar to that of E. coli. Synthesis is bidirectional and continuous on one strand and discontinuous on the other, and the requirements of synthesis (four deoxyribonucleoside triphosphate, divalent cation, template, and primer) are the same. Okazaki fragments of eukaryotes are about one-tenth the size of those in bacteria. Because there is a much greater amount of DNA to be replicated and DNA replication is slower, there are multiple initiation sites for replication in eukaryotes (and increased DNA polymerase per cell) in contrast to the single replication origin in bacteria. Replication occurs at different sites during different intervals of the S phase.

Distinguish between (a) unidirectional and bidirectional synthesis, and (b) continuous and discontinuous synthesis of DNA.

Given a stretch of double-stranded DNA, one could initiate synthesis at a given point and either replicate strands in one direction only (unidirectional) or in both directions (bidi- rectional) Notice the synthesis of complementary strands occurs in a continuous 5′ to 3′ mode on the leading-strand template in the direction of the replication fork, resulting in a single, long product strand, and in a discontinuous 5′ to 3′ mode on the lagging strand resulting in shorter product strands called Okazaki fragments.

Regarding the figure in the question above, what is the difference between the observed results for lanes 2 (medium DNase amount) and 3 (high DNase amount)?

Medium levels of DNase digest between every nucleosome, while high levels completely digest the linker DNA.

If an inversion moved a gene from one region of euchromatin to another region that was also euchromatin, would would you expect to be most likely?

Minor or no effect on the gene's expression

About 2% of all people have inversions that are noticeable by karyotype, yet the vast majority are of no phenotypic consequence. Why is this the case?

Most inversions do not disrupt a gene.

Why were 32P and 35S chosen for use in the Hershey-Chase experiment? Discuss the rationale and conclusions of this experiment.

Nucleic acids contain large amounts of phosphorus and no sulfur, whereas proteins contain sulfur and no phosphorus. Therefore, the radioisotopes 32P and 35S will selectively label nucleic acids and proteins, respectively. The Hershey and Chase experiment was based on the premise that the substance injected into the bacterium is the substance responsible for producing the progeny phage and therefore must be the hereditary material. The experiment demonstrated that most of the 32P labeled material (DNA) was injected, while the 35S labeled phage ghosts (protein coats) remained outside the bacterium. Therefore, the nucleic acid must be the genetic material.

If DNA polymerase could synthesize DNA in both the 5'->3' and 3'->5' directions, what would not be needed?

Okazaki fragments

What is the expected phenotype of a telomerase mutant (assume loss of function)?

chromosomes will become shorter and shorter each generation

In density gradient centrifugation of Okazaki fragments and the leading strand, where would you expect to find the leading strand?

closer to the bottom of the centrifuge tube (i.e. below the Okazaki fragments)

If resistance arose in response to antibiotic treatment, what pattern would you expect if you replica printed from a plate without antibiotics to multiple plates with antibiotics?

colonies would be randomly located when comparing antibiotic plates

Which type of abnormality is most likely to cause fertility issues?

heterozygous translocation

Thermophiles are microbes that grow at high temperatures (>70°C). What are your expectations of their GC content relative to organisms that grow at moderate temperatures?

higher percentage of GC pairs

Unequal crossing-over is more likely to occur at . . .

highly similar sequences

The basic amino acids of histones serve to . . .

increase histone affinity for DNA

Rank the following inversions in by most to least likely to alter a phenotype?

inversion breakpoint occurs within a gene -> gene repositioned near heterochromatin -> gene repositioned near euchromatin

If mutations were specifically induced as a direct response to selective pressure, what would have been the result of Esther and Joshua Lederberg's replica printing experiment? (Reminder: the experimental design was to isolate colonies on a plate grown without antibiotic, and then replica print to multiple plates with antibiotic).

Resistant colonies would appear to be randomly distributed across the different antibiotic plates.

At the end of the short arm of human chromosome 16 (16p), several genes associated with disease are present, including thalassemia and polycystic kidney disease. When that region of chromosome 16 was sequenced, gene-coding regions were found to be very close to the telomere associated sequences. Could there be a possible link between the location of these genes and the presence of the telomere associated sequences? What further information concerning the disease genes would be useful in your analysis?

Since both genes mentioned in the problem are near the telomeric heterochromatin, they may be subject to silencing due to the position effect. It is also possible that erosion of the end of the chromosome is related to each disease. Examination of the gene by in situ hybridization and molecular cloning indicates that thalassemia involves a terminal deletion in the distal portion of 16p

What is a plausible explanation for the fact that trisomy 21 is the only human aneuploidy with survival past infancy?

Smaller chromosomes decrease the likelihood that gene dosage imbalance will be lethal.

In 1994, telomerase activity was discovered in human cancer cell lines. Although telomerase is not active in human somatic tissue, human somatic cells do contain the genes for telomerase proteins and telomerase RNA. Since inappropriate activation of telom- erase may contribute to cancer, why do you think the genes cod- ing for this enzyme have been maintained in the human genome throughout evolution? Are there any types of human body cells where telomerase activation would be advantageous or even necessary? Explain.

Telomerase activity is present in germ-line tissue to maintain telomere length from one generation to the next. It is also necessary in stem cells and other proliferating tissues. In other words, telomeres cannot shorten indefinitely without eventually eroding genetic information.

Describe the role of 15N in the Meselson-Stahl experiment.

The Meselson and Stahl experiment has the following com- ponents: By labeling the pool of nitrogenous bases of the DNA of E. coli with heavy isotope 15N, it would be possible to "follow" the "old" DNA. Cells were grown for many generations in medium containing 15N and then transferred to 14N medium so that "new" DNA could be detected. A comparison of the density of DNA samples at various times in the experiment (initial 15N culture and subsequent cultures grown in the 14N medium) showed that after one round of replication in the 14N medium, the DNA was half as dense (intermediate) as the DNA from bacteria grown only in the 15N medium. In a sample taken after two rounds of replication in the 14N medium, half of the DNA was of the intermediate density and the other half was as dense as DNA containing only 14N DNA.

You perform the Ames test on a newly discovered antibiotic to determine whether it is safe for human use. The results are below: (sample : number of His+ revertant colonies) Water: 10 Water + rat liver enzymes: 7 Drug: 150 Drug + rat liver enzymes: 8 What is your interpretation of the data?

The drug is mutagenic, but rat liver enzymes convert it into non-mutagenic products

True or False: Eukaryotic DNA is also supercoiled.

True

Inversions are said to "suppress crossing over." Is this terminology technically correct? If not, restate the description accurately.

While there is the appearance that crossing over is sup- pressed in inversion "heterozygotes," the phenomenon extends from the fact that the crossover chromatids end up being abnormal in genetic content. As such, they fail to produce viable (or competitive) gametes, or they lead to zygotic or embryonic death.

Does the data from the Luria-Delbruck experiment support the hypothesis that mutations are random or that they are a response to the environment?

random

If DNA polymerase could synthesize DNA in both the 5' to 3' direction and 3' to 5' direction, what would not be needed?

telomerase

Seedless varieties of fruit have been developed by producing plants with abnormal euploidy, which causes abnormal meiosis and thus few seeds. Which of the following ploidies would you expect to be common in seedless varieties?

triploidy

Facioscapulohumeral disease (FSDH) is an autosomal dominant neuormuscular disorder, characterized by progressive weakness of facial, shoulder, and upper arm muscles. FSDH is caused by a deletion of repetitive DNA that brings the gene responsible for the disorder closer to the telomere, leading to variegated expression. Because of this variability in expression, this disorder shows...

Incomplete Penetrance

If protein was the genetic material of bacteriophage, where would you expect the 35S radiolabel to end up?

Infected bacteria, phage ghosts, & viable phages

If the transforming principle was a protein, which treatment would result in no transformation?

Protease

For Barbara McClintock's Ac/Ds system, let's assume the Ds element is currently inserted into the c gene, leading to the white (mutant) kernel phenotype. What is the expected phenotype for the kernels if you deleted only the inverted repeat sequences of the Ac element?

Purple spots would randomly appear on the kernels.

A couple planning their family are aware that through the past three generations on the husband's side a substantial number of stillbirths have occurred and several malformed babies were born who died early in childhood. The wife has studied genet- ics and urges her husband to visit a genetic counseling clinic, where a complete karyotype-banding analysis is performed. Although the tests show that he has a normal complement of 46 chromosomes, banding analysis reveals that one member of the chromosome 1 pair (in group A) contains an inversion covering 70 percent of its length. The homolog of chromosome 1 and all other chromosomes show the normal banding sequence. (a) How would you explain the high incidence of past stillbirths? (b) What can you predict about the probability of abnormality/ normality of their future children (c) Would you advise the woman that she will have to bring each pregnancy to term to determine whether the fetus is normal? If not, what else can you suggest?

a) In all probability, crossing over in the inversion loop of an inversion (in the heterozygous state) had produced defective, unbalanced chromatids, thus leading to stillbirths and/or malformed children. b) It is probable that a significant proportion (perhaps 50 percent) of the children of the man will be similarly influenced by the inversion. c) Since the karyotypic abnormality is observable, it may be possible to detect some of the abnormal

Several temperature-sensitive mutant strains of E. coli display the following characteristics. Predict what enzyme or function is being affected by each mutation. a) Newly synthesized DNA contains many mismatched base pairs. b) Okazaki fragments accumulate, and DNA synthesis is never completed. c) No initiation occurs. d) Synthesis is very slow. e) Supercoiled strands remain after replication, which is never completed.

a) No repair from DNA polymerase I and/or DNA poly- merase III b) No DNA ligase activity c) No primase activity d) Only DNA polymerase I activity e) No DNA gyrase activity

Define and indicate the significance of (a) Okazaki fragments, (b) DNA ligase, and (c) primer RNA during DNA replication.

a) Okazaki fragments are relatively short (1000 to 2000 bases in bacteria) DNA fragments that are synthesized in a discontinuous fashion on the lagging-strand templates during DNA replication. Such fragments appear to be necessary because template DNA is not available for 5′ to 3′ synthesis until some degree of continuous DNA synthesis occurs on the leading- strand template in the direction of the replication fork. The isolation of such fragments provides support for the scheme of replication (b) DNA ligase is required to form phosphodiester linkages in nicks, which are generated when DNA polymerase I removes RNA primer and meets newly synthesized DNA ahead of it. The discontinuous DNA strands are ligated together into a single continuous strand. c) Primer RNA is formed by RNA primase to serve as an initiation point for the production of DNA strands on a DNA template. None of the DNA polymerases are capable of initiating synthesis without a free 3′ hydroxyl group. The primer RNA provides that group and thus can be used by DNA polymerase III.

Assume that the Ds element is currently inserted in the the c gene, leading to white (mutant phenotype) kernels. What phenotype would you expect if you also mutated the transposase gene of the Ac element, making it non-functional?

all kernels would remain white (mutant)

The figure below represents homologous chromosome pairing during meiosis. Which type of mutation could cause a loop structure to form?

duplication and deletion

When yeast cells are evolved by growing them continuously on low glucose media, mutants arise with a growth advantage. What types of mutations would you expect to give a growth advantage?

duplications of glucose transporter genes

Most E. coli strains are unable to grow on citric acid as a source of carbon and energy. However, a small number of strains are naturally able to grow on citric acid. In these strains, where would you expect to find the gene(s) that allow for this novel trait of citric acid metabolism?

on a plasmid

Where would you expect to find the gene that encodes ligase in E. coli?

on the chromosome