Biology 201 (Genetics)

Name ______

Biology 201 (Genetics) Exam #3

 Read the question carefully before answering. Think before you write. Be concise.

 You will have up to one hour to take this exam. After that, you MUST stop no matter where you are in the exam.

 If I can not read your handwriting, I will count the question wrong.

 Sign the honor pledge if applicable.

 Good luck!

I pledge that I have neither given nor received unauthorized assistance during the completion of this work. Signature: ______

1) Complete the following table. Indicate the polarity at the left end of each row. Identify the template DNA strand for transcription with an arrow.

5’ A T G T G G C G T T G A DNA double 3’ T A C A C C G C A A C T helix  5’ A U G U G G C G U U G A mRNA 3’ U A C A C C G C A A C U tRNA anticodon MET Trp ARG Stop Amino acids 2) Missense point mutations are isolated in the genes for E. coli RNA polymerase core subunits much more frequently than frameshift mutations. Why do you think this is?

Missense point mutation are mutations that change one basepair and thus at the most change only one amino acid in the resulting protein. This may or may not have an effect on the function on the protein depending upon how important the chnaged amino acid is for the proper function of the protein. Frameshift mutations are mutations that insert one or two basepairs into the nucleotide sequence, thereby changing the entire reading frame after the insertion. This results in a protein that is nonfunctional since all the amino acids after the point of insertion will be incorrect. Thus, since RNA polymerase genes are certainly essential for the cell to live (since transcription is essential for life), the reason that missense mutations are isolated more frequently than frameshift mutations is that the frameshift mutants are all dead!

3) You put your yeast plates in the dark after UV irradiation in lab. Why did you do this?

To prevent repair of the UV induced thymine-thymine dimmers in the DNA by photoreactivation.

How would your UV survival results have been different if you let plates incubate in the light? Why?

You would have seen an increased survival because of photoreactivation DNA repair. 4) In a certain community, a widespread mutant allele for Tay–Sachs disease (autosomal recessive) has a mutated HindIII restriction enzyme site within the gene so that it is no longer cut by HindIII. A DNA probe is available for part of the gene, and this probe binds to both the wildtype and mutant alleles. The regions of homology with the probe are indicated with the shaded boxes.

HindIII HindIII HindIII Wildtype allele:

HindIII HindIII Tay-Sachs mutant allele:

A couple in the community is expecting a child and they have DNA tests to determine whether the child will have Tay-Sachs. DNA from the parents and the fetus is obtained, digested with HindIII, and ran on an agarose gel. A Southern blot is then done with the probe described above. The results of the genetic tests are shown below: mom dad fetus

a)On the gel, identify the bands that correspond to the Tay-Sachs allele with a lowercase t and the bands that correspond to the normal allele with an uppercase T.

b)How would you council the couple in regards to the genotype and phenotype of the fetus that the mother is carrying? Genotype is tt; phenotype is Tay-Sachs

c)How would you council the couple in regards to their genetic status for future pregnancies? They are both heterozygotes and so with each pregnancy there is a ¼ chance that they would have a child with Tay-Sachs. 5) 5) A Southern blot from a single VNTR locus probe analysis of various DNA samples in a rape investigation is shown in the figure. The DNA samples are as follows: V = victim, A = suspect A, B = suspect B, and E = evidence from the crime scene.

a) If you are the DNA analyst, what should you conclude about the possibility of suspect A being guilty of the crime? His VNTR pattern for this locus matches the evidence so he is possibly guilty of the crime.

b) If you are the DNA analyst, what should you conclude about the possibility of suspect B being guilty of the crime? His VNTR pattern for this locus does not match the evidence so he not guilty of the crime.

c) What would you do to make this a more conclusive genetic forensics test? Find out the frequencies of the alleles of suspect A in the general population so you can acertain the probability that this is essentially a unique VNTR pattern. Also, probe with several other VNTR locus probes to decrease the probability that the match is from another person.

Blood samples

6) Choose ONE of the two following questions to answer. Only answer one of the two. If you answer both, I will only grade the first one. Answer 6A OR 6B. You can word your answer in list format if you want (i.e. complete sentences are not needed).

OPTION 6A: A process for the large scale production of human growth hormone (HGH) would be of great use for therapeutic applications in the treatment of dwarfism, increase of bone density and anti-aging. Thus, you want to clone the human growth hormone gene (hgh) and produce the protein in E. coli. Briefly explain the best way to do this. You have the following tools available to you: a DNA probe of the growth hormone gene from monkeys, antibodies to HGH protein, any vector you would like, human cells, and E. coli. Please note that you do NOT have to use all of these reagents in your answer. You can word your answer in list format if you want (i.e. complete sentences are not needed).

OPTION 6B: The gene for the protein VitB which allows an organism to synthesize its own vitamin B has been cloned from the yeast Yeastie beastie. The gene is called vitB. You are a research scientist studying vitamin B synthesis in the closely related fungus Fungus amongus. You wish to clone the vitB gene from F. amongus. There are a variety of different ways that you could do this. Briefly, describe one of the ways that you could clone the F. amongus vitB gene. You have the following tools available to you: a DNA probe of the vitB gene from Yeastie beastie, antibodies to VitB protein, a mutant in F. amongus that can not grow unless you supply vitamin B in the medium. Please note that you do NOT have to use all of these tools in your answer; in fact, one tool should suffice. You can word your answer in list format if you want (i.e. complete sentences are not needed).

6A: Method #1: Construct a plasmid library containing human cDNA  Screen the library for the hgh gene with the cloned gene from monkey. Construction of the library: 1. Make cDNA from human cell mRNA using reverse transcriptase 2. Isolate plasmid DNA for vector from bacteria. 3. Digest cDNA and plasmid DNA with the same restriction enzymes. 4. Ligate the sticky ends of the restriction enzyme digested cDNA from human cells and plasmid DNA together using DNA ligase. Transform the recombinant DNA molecules into E. coli for amplification of the DNA. Plate the transformed DNA onto agar plates containing the antibiotic to which the plasmid confers resistance. Each colony came from one cell that contains a recombinant plasmid that contains one fragment of human cDNA.

Screening the library for the colony that contains human hgh gene using the hgh gene from monkey as a probe: 1. Transfer the colonies to filter paper. 2. Obtain the radioactive hgh gene from monkey. Denature by heating so that it is single standed. 3. Place the filter into a bag along with the radioactive hgh gene from monkey. 4. Because the radioactive hgh gene from monkey is homologous to the hgh gene from human, they will bind to each other. This results in a radioactive spot on the filter corresponding to the colony that contains the plasmid carrying the human cDNA fragment that has the hgh gene. 5. Detect the radioactive spot using autoradiography (exposing a piece of X-ray film). 6. Pick the colony that corresponds to the radioactive spot. 7. Inoculate fresh media with the colony. The bacteria will grow, thereby amplifying the plasmid that contains the hgh gene.

Method #2: Construct a plasmid DNA library containing human cDNA  Screen the library for the hgh gene with the antibodies to Hgh. Construction of the library: As per above example.

Screening the library for the colony that contains human hgh gene using Hgh antibodies. 1. Transfer the colonies to filter paper. 2. Obtain the antibody to Hgh. 3. Place the filter into a bag along with the antibody to Hgh. 4. The Hgh antibody will bind to the Hgh made by the colonies that contain the plasmid with the hgh gene. 5. Add a radioactive secondary antibody that binds to the primary antibody in step 4. 6. Detect the radioactive spot using autoradiography (exposing a piece of X-ray film). 7. Pick the colony that corresponds to the radioactive spot. 8. Inoculate fresh media with the colony. The bacteria will grow, thereby amplifying the plasmid that contains the hgh gene.

Method 3: Use a Southern blot to ID the cDNA fragment with the hgh gene and then clone into plasmid as described above in the first section. This answer received almost all the possible points. The only thing wrong with this approach was that even though you can ID a band on a Southern blot as being the gene of interest, it is not truly isolated from all the other DNA fragments. Remember when you cut the entire chromosome of an organism with a restriction enzyme, there are thousands of bands, some of which may be very similar in size. So you can not just do a Southern blot and then cut out the band of the correct size for cloning because there will be other bands of the same size with different genes at the same place on the blot/gel. You still have to screen your clones for the correct gene as mentioned above. 6B: Method #1: Construct a plasmid DNA library containing F. amongus DNA  Screen the library for the vitB gene with the cloned gene from Y. beastie. Construction of the library: 1. Isolate chromosomal DNA from F. amongus and plasmid DNA for vector from bacteria. 2. Digest chromosomal DNA from F. amongus and plasmid DNA with the same restriction enzymes. 3. Ligate the sticky ends of the restriction enzyme digested chromosomal DNA from F. amongus and plasmid DNA together using DNA ligase. Transform the recombinant DNA molecules into E. coli for amplification of the DNA. Plate the transformed DNA onto agar plates containing the antibiotic to which the plasmid confers resistance. Each colony came from one cell that contains a recombinant plasmid that contains one fragment of DNA from the F. amongus chromosome.

Screening the library for the colony that contains the F. amongus vitB gene using the vitB gene from Y. beastie as a probe: 1. Transfer the colonies to filter paper. 2. Obtain the radioactive vitB gene from Y. beastie. Denature by heating so that it is single standed. 3. Place the filter into a bag along with the radioactive vitB gene from Y. beastie. 4. Because the radioactive vitB gene from Y. beastie is homologous to the vitB gene from F. amongus, they will hybridize to each other. This results in a radioactive spot on the filter corresponding to the colony that contains the plasmid carrying the DNA fragment from F. amongus that has the vitB gene. 5. Detect the radioactive spot using autoradiography (exposing a piece of X-ray film). 6. Pick the colony that corresponds to the radioactive spot. 7. Inoculate fresh media with the colony. The bacteria will grow, thereby amplifyiing the plasmid that contains the vitB gene.

Method #2: Construct a plasmid DNA library containing F. amongus DNA  Screen the library for the vitB gene with the antibodies to VitB. Construction of the library: As per above example except the library should be made from cDNA (DNA made from mRNA via reverse transcriptase) so that there are no introns in the gene.

Screening the library for the colony that contains the F. amongus vitB gene using VitB antibodies. 1. Transfer the colonies to filter paper. 2. Obtain the antibody to VitB. 3. Place the filter into a bag along with the antibody to VitB. 4. The VitB antibody will bind to the VitB made by the colonies that contain the plasmid with the vitB DNA fragment. 5. Add a radioactive secondary antibody that binds to the primary antibody in step 4. 6. Detect the radioactive spot using autoradiography (exposing a piece of X-ray film). 7. Pick the colony that corresponds to the radioactive spot. 8. Inoculate fresh media with the colony. The bacteria will grow, thereby amplifying the plasmid that contains the vitB gene.

Method 3: Use a Southern blot to ID the fragment with the hgh gene and then clone into plasmid as described above in the first section. This answer received almost all the possible points. The only thing wrong with this approach was that even though you can ID a band on a Southern blot as being the gene of interest, it is not truly isolated from all the other DNA fragments. Remember when you cut the entire chromosome of an organism with a restriction enzyme, there are thousands of bands, some of which may be very similar in size. So you can not just do a Southern blot and then cut out the band of the correct size for cloning because there will be other bands of the same size with different genes at the same place on the blot/gel. You still have to screen your clones for the correct gene as mentioned above. 7)The elements responsible for directing RNA polymerase to a particular gene to begin transcription include a. initiation factors and rho b. promoter, sigma factor, and rho c. promoter, sigma factors, and elongation factors d. promoter and initiation factors d. promoter and sigma factors

8)More than one codon can specify the amino acid serine because a. there is more that one tRNA type that carry serine and that have different anticodons b. the serine tRNA may wobble when binding to a codon and therefore recognize more that one codon c. there is more than one serine tRNA synthetase to put serine on different tRNAs d. a and b e. all of the above

9)Tetracyline inhibits binding of charged tRNAs to bacterial ribosomes, except fmet-tRNA. The most likely target of tetracyline is: a. the anticodons b. the ribosome binding site c. aminoacyl tRNA synthetases d. the A site of the ribosome e. the ribosomal RNA (rRNA)

10) Cycloheximide inhibits peptidyl transferase activity of the ribosome. If cycloheximine were added to an actively translating ribosome, one would observe a stalled (non- moving) ribosome with: a. two tRNAs, one carrying a long polypeptide chain, another carrying a single amino acid. b. two tRNAs, neither carrying any amino acids. c. two tRNAs, each carrying polypeptide chain of equal lengths d. one tRNA in the P site and holding a polypetide chain. e. one tRNA in the A site and holding a polypetide chain.

11) What would be the effect if you inhibited splicing of eukaryotic mRNA? a. mRNA would not be polyadenylated and therefore would be degraded  no protein made b. mRNA would not be capped and thus would remain in the nucleus  no protein made c. mRNA would contain both introns and exons  proteins made containing extra amino acids that are normally not present d. mRNA that contained only introns  proteins made containing the wrong amino acids e. mRNA that contained only exons  proteins made containing the wrong amino acids BONUS : The next CAT that adopts me is going to be named after the amino acid histidine. Provide a genetic rationale for why I would name my CAT histidine.

Because the DNA sequence on the nontemplate strand that codes for the amino acid histidine is CAT.