Exam 2 answers 2001

1) 28 pt. Answer True or False. If the statement is TRUE, explain or give a supporting example. If the statement is FALSE, either explain the problem with the statement or give a counter example. If you think the answer is ambiguous, state your case.

(a) A promoter down mutation--one that binds RNAP less well--would likely be a hypomorph.

True. A promoter down mutation should decrease expression but not eliminate it. This is the definition of a partial loss-of-function mutation.

(b) A promoter up mutation--one that binds RNAP better--would likely be a neomorph.

Most likely false. An up mutation would lead to over expression of the wild-type protein and be a hypermorph. A neomorphic mutation could arise if the unregulated promoter expressed the protein in cells or at times where and when it normally did not.

(c) A point mutation within an intron that prevented inclusion of a small 15 bp exon in the middle of a gene would most likely be classified as an amorphic allele.

False. A 15 bp deletion would keep the appropriate reading frame, and its quite likely the protein would still function minus the 5 aa stretch. In that case it would be a hypomorph. If the 5 aa's were essential the mutation would probably be an amorph (null).

(d) A tRNA suppressor would likely be classified as antimorphic.

A tRNA suppressor is probably best described as neomorphic. It has acquired a function the wild-type gene does not have, and the mutations are dominant.

(e) Post-replication repair exploits DNA methylation to determine which damaged strand to fix.

True. The older, template strand is fully methylated and the post-replication repair pathway exploits this to repair the newly synthesized strand.

(f) The E. coli ada repair system is an example of an excision repair pathway.

False. The ada system is a mechanism for direct repair. The lesion is removed directly from the damaged base.

(g) The following enzymes are needed to make a cDNA library: DNA polymerase, DNA ligase, RNA polymerase, restriction enzymes.

False. All the enzymes are needed except RNA polymerase. Two kinds of DNA polymerase are used. Reverse transcriptase (to copy the mRNA) and E. coli or phage T4 DNA polymerase for synthesis of the other strand. (Full credit for RT.)

2) 15 pt. The following event, steps, or reactions occur during E. coli replication. For each entry in column A, select the appropriate entry in column B. Each entry in A may have more than one answer and each entry in B can be used more than once or not at all. If you think any of your answers are ambiguous, briefly explain why.

	Column A	Column B
C (G is allowable)	a) Unwinds the double helix	A Polymerase I
D	b) Is an RNA polymerase	B Polymerase III
A B	c) Is a DNA polymerase	C) helicase
B	d) is the major elongation enzyme	D) Primase
A B D	e) A 5' to 3' polymerase	E) Ligase
H	f) A 3' to 5' polymerase	F) SSB protein
A B	g) A 3' to 5' exonuclease	G) Topoisomerase
A	h) A 5' to 3' exonuclease	H) None of the above
H (G)	i) A sequence non-specifc endonuclease
E	j) Bonds 3'-OH of a polynucleotide to a free 5'-phosphate of a polynucleotide
F	k) Binds to single-stranded DNA

3) 16 pt. A functional human hemoglobin variant called Constant Spring, is known to occur at high frequency in certain Asian populations. Normal a-hemoglobin is 141 amino acids long, while the Constant Spring a polypeptide is 172 amino acids long.

a) Explain how the Constant Spring allele could have arisen via a single-base substitution or by a frameshift mutation. (Answer for both.)

A frameshift mutation late in the coding sequences would change the frame of the normal termination codon and translation would continue until it reached a stop codon in the new reading frame.
A single base substitution that changed the normal stop codon to a sense codon would result in a longer protein since translation would continue beyond the normal stop site until a new nonsense codon was reached.

Beginning with the last codon (Arg) of the wild-type a-globin gene, the 3' end of the wild-type a-globin mRNA has the sequence:

5'..CGU UAA GCU GGA GCC UCG GUA GCA GUU CCU CCU GAC AGA UGG GCC UCC CAA CGG GCC CUC CUC CCC UCC UUG CAC CGG CCC UUC CUG GUC UUU GAA UAA AGU CUG AGU GGG....3'.

b) Based on this information how was the Constant Spring allele likely generated? Support your answer. What are the last three amino acids of the mutant a-hemoglobin?

Constant Spring hemoglobin is caused by a single base change in the termination codon UAA of the a-globin gene. This mutation must generate a sense codon. Such a change would lead to a 31 amino acid extension of the a-globin polypeptide since the next in frame stop codon (UAA) is located (31 X 3) bases 3' of the normal stop codon. The last three amino acids are V (GUC) F (UUU) and E (GAA).

Neither a +1 (-2) or -1 (+2) frameshift would produce the Constant Spring phenotype. A + 1 frameshift late in the gene would lead to a 10 amino acid extension because translation would terminate at the underlined UGA sequence. A - 1 frameshift would result in a 5 amino acid extension because translation would terminate at the underlined UAG sequence.

4) 12 pt. The first child of a couple was born with an autosomal recessive disease causing severe mental retardation. They would like to have another child soon but want to ensure it will not inherit the disease. The normal version of the gene that can be mutated to cause the disease is known. In fact, it has been mapped and sequenced as part of the human genome project. Unfortunately, the nature of the disease causing allele(s) is (are) unknown. Suggest an approach that could be used to help this family.

Use the human genome sequence to find micro- or mini-satellite sequences that are closely linked to the structural gene. Such sequences are very common in the human genome and numerous examples of each would likely be found within a relatively short distance ( in terms of cM's) of the gene. Examine the parents to determine if they are polymorphic for any of these loci. If so, one can determine which alleles are linked to the disease causing mutation by comparing the micro- or minisatellite alleles present in the affected child and the two parents. If such polymorphic sequences can be found, the allelic state of the new fetus can be determined by sampling cells from the amniotic fluid. From that information one can calculate the probability of the fetus having the disease causing genotype.

If the couple was opposed to abortion of the fetus, the same techniques could be applied to embryos created in vitro. A few cells from a developing embryo could be genotyped for the polymorphic loci, and only embryos with the non-mutant genotype could be implanted into the mother.

5) 16 pt. In a cross designed to determine the frequency of cotransduction between fol and pyrA, you use as a donor, a strain that is fol-, pyrA+, and as a recipient, a strain that is fol+ pyrA-. pyrA- strains require a pyrimidine and arginine to grow; fol- strains require folic acid to grow but they are also resistant to aminopterin. How would you carry out the cross and score transductants? Indicate media components assuming you have a basal medium that contains all the salts and trace minerals necessary to support the growth of a prototrophic strain.

One method.

1) Grow phage P1 on fol- pyrA+ donor.

2) Infect fol+ pyrA- recipient with lysate.

3) Plate mix on minimal glucose plates + folic acid to select for pyrA+.

4) Score the pyrA+ transductants for fol phenotypes by plating on minimal glucose + folic acid + or - aminopterin. fol- will grow on both plates, fol+ cells will not grow with aminopterin.

5) % cotransduction = #aminopterin resistant/total pyrA+ X 100.

variation:

4b) score by plating on minimal glucose + or - folate. fol- will grow only on the supplemented plate.

5b) % cotransduction = # growing only on + folate plate /total pyrA+ X 100.

A second method.

1) Grow phage P1 on fol- pyrA+ donor.

2) Infect fol+ pyrA- recipient with lysate.

3) Plate mix on minimal glucose plates + uracil + arginine + folate + aminopterin to select for fol- transductants..

4) Score the fol- transductants for pyrA phenotype by plating on minimal glucose + folate and minimal glucose + folate + uracil + arginine. pyrA+ will grow on both plates, pyrA- cells will grow only on the secnd plate.

5) % cotransduction = # pyrA+ / total fol- X 100.

Now reverse the donor and recipient strains. How would you carry out the cross, and score the transductants?

1) Grow phage P1 on fol+ pyrA- donor.

2) Infect fol- pyrA+ recipient with lysate.

3) Plate mix on minimal glucose plates + uracil + arginine to select for fol+ transductants.

4) Score the fol+ transductants for pyrA phenotype by plating on minimal glucose and minimal glucose + uracil + arginine. pyrA+ will grow on both plates, pyrA- cells will only grow on the supplemented plate.

5) % cotransduction = # pyrA- / total fol+ X 100.

6) 14 pt. A particular Drosophila strain carries a mutation that conditionally overexpresses a growth factor. The phenotype caused by the big gross maggot, bgm, allele is that the larvae grow to giant size. The Japanese film industry is planning to shoot a series of new Mothra v. Godzilla movies. Their plan is to feed large quantities of the silk moth bgm protein to larvae to create giant silk moths for use in live-action shots in order to save on the costs of computer animation. Your work on bgm has drawn the attention of the movie producers and they have given you a grant to clone the silk moth bgm gene and overproduce the protein.

Starting with the sequenced Drosophila bgm genomic clone, outline the steps you would use to accomplish the task of overproducing the silk moth bgm protein. Be sure to state any special experimental conditions you need. You may assume silk moth genomic and cDNA libraries exist.

Overview:

Radiolabel a Drosophila genomic bgm exon probe and use it to screen a silk moth cDNA library under conditions of low- or moderate-stringency. Isolate the silk moth bgm cDNA and clone it into an expression vector designed to produce protein.

Details.

The exon sequences should be used as a probe as they are most likely to be conserved between the species. An intron probe could easily lead to false positives. PCR amplify or cut the probe from the Drosophila clone with a restriction enzyme and label it using random primer labeling.

Plate the silk moth cDNA library (there is no need to use a genomic silk moth library in this experiment), transfer the phage to a filter and incubate with the ³²P-probe under appropriate low-stringency conditions, i.e., reduced temperature and high salt until a "reasonable" number of positive signals result. Isolate the positive plaques, by repeating the hybridization procedure until a pure population of cDNA bearing phage is obtained.

Sequence the silk moth cDNA inserts to identify those similar in sequence to the fly bgm coding sequences

For the positives, clone the s.m. bgm cDNA into an expression vector that will allow the production of large amounts of s.m. bgm protein in E. coli. The cDNA must contain the entire coding sequence of the s.m. protein from the initiating ATG to the termination codon. This would likely most easily be done by PCR amplifying the coding region of the cDNA and ligating it into the overexpression vector.

Extra credit —6pt.

The Avery, and Hershey and Chase experiments proved that DNA was the hereditary material in bacteria and in DNA-containing viruses. Some viruses have RNA and no DNA inside the viral particle. An example is tobacco mosaic virus TMV. TMV infects tobacco plants causing spotted lesions. Two different strains of TMV have different forms of a viral protein that can be distinguished by whether they cause large or small spots on the tobacco leaves.

It is possible to reconstitute TMV in the test tube by mixing pure viral proteins and pure viral RNA. The reconstituted virus can be used to infect host plants to create a new generation of virus. Design an experiment to show that RNA acts as the hereditary material in TMV.

Create hybrid viruses containing different protein coats and RNA strands. For example, reconstitute a virus with large spot protein + small spot RNA. Use this to infect a plant, where it will grow and produce progeny. The genetic material should dictate the phenotype of those progeny. If RNA is the genetic material, the progeny virus should produce small spots when used to infect a new plant, and the virus should be true breeding in subsequent generations. If protein were the genetic material, the progeny virus would produce large spots.

One should also do the reciprocal experiment to create a small spot protein + large spot RNA virus. In this case, the progeny virus should create large spots and be thereafter true breeding if RNA is the genetic material.