Problem Set 2, BioS 220, 2011: DNA Replication, Transcription, and Translation, Study notes of Genetics

Download Problem Set 2, BioS 220, 2011: DNA Replication, Transcription, and Translation and more Study notes Genetics in PDF only on Docsity! Problem set 2, BioS 220, 2011 Name _________________________ TA ______________________ 1 1. If during DNA replication, RNA bases are found in the newly synthesized DNA, which enzyme or enzymes may be defective? (1 point) DNA polymerase I. DNA polymerase I has 5’ to 3’ exonuclease activity and normally cleaves RNA primers to replace with DNA. 2. In the fruit fly Drosophila melanogaster, DNA replication at a single replication fork occurs at a rate of about 2600 nucleotide pairs per minute. The DNA molecule occurring in one of the largest chromosomes of this species has been estimated to contain 6 x 10 7 nucleotide pairs. If replication of this molecule was initiated at a single origin in the middle of the chromosome, estimate the time, in days, required for complete replication of the chromosome. (1 point) 6 x 10 7 basepairs / (2 forks per origin x 2600 basepairs per fork per minute) = 11538 minutes = 8 days 3. The following figure shows a replication fork in a segment of DNA, You are viewing the part that is just opening. (2 points) a. Along which template strand (upper or lower) does continuous synthesis occur? (1 point) Lower b. Along which template strand are numerous Okazaki fragments formed? (1 point) Upper 4. The temperature at which half of the hydrogen bonds in a sample of duplex DNA have denatured is called the melting point, or Tm. Listed below are the base compositions of three different duplex DNA molecules. I. 5’ ATTCTAACTTTGAT 3’ II. 5’ CGCACTGGCTAACC 3’ III. 5’ CGCATTATTGTCAA 3’ 3’ TAAGATTGAAACTA 5’ 3’ GCGTGACCGATTGG 5’ 3’ GCGTAATAACAGTT 5’ Order these three molecules in terms of their melting points from highest to lowest. (1 point) II > III > I II has 37 hydrogen bonds; III has 33 hydrogen bonds; I has 31 hydrogen bonds 3’ 5’ replication fork moves this way Problem set 2, BioS 220, 2011 Name _________________________ TA ______________________ 2 5. Label the diagram below to show a site where each of the following enzymes could act during DNA replication: DNA polymerase III, DNA ligase, helicase, topoisomerase. Label the polarity of both template strands. (5 points) 6. Below is a graph showing DNA content against time in a cell that undergoes mitosis. Assuming it is a diploid cell, outline the period of S phase. (1 point) 7. True of False: (5 points total) Ta ) DNA replication in eukaryotes only occurs during the S phase F b) During DNA replication, the role of the sliding clamp is to rewind the DNA T c) If the DNA in my liver has a GC content of ~30%, the DNA in my heart also has a GC content of ~30% T d) During DNA replication, the RNA primers are likely to contain errors because primase lacks a proofreading function F e) At the replication fork, one daughter strand grows from 5’ to 3’, and the other daughter strand grows from 3’ to 5’ Problem set 2, BioS 220, 2011 Name _________________________ TA ______________________ 5 17. A double-stranded DNA molecule with the sequence shown below produces, in vivo, a polypeptide that is 7 amino acids long. (1 point) 5’-TAC ATG ATC ATT TCA CGG AAT TTC TAG CAT GTA-3’ 3’-ATG TAC TAG AGT TAA GCC TTA AAG ATC GTA CAT-5’ Which strand (upper or lower) of the DNA is the coding strand? The upper strand 18. Drug X disrupts protein synthesis. Suppose you are doing a translation reaction in vitro and and you add drug X just after the 2nd bond is made. You wait a few minutes and look at the ribosome. You see that the A site is empty. Which step of translation is likely inhibited by drug X? Why? (2 points) Drug X prevents the ternary complex from entering the A site. (Also right: Drug X prevents the aminoacyl-tRNA from binding to the A site.) 19. Mutation in which of the following sequences may abolish protein translation, but not gene transcription, in prokaryotes? (1 point) a) The Shine-Dalgarno sequence b) The TATA box c) The -35 box d) All of the above 20. The peptidyl-transferase center is composed of ___ _rRNA_____________ (1 point) 21. There are 6 codons for the amino acid Serine; different tRNAs are needed to service these codons. Based on the codon table above and the wobble rule (below), what is the minimum number of Serine tRNAs needed and what are their respective anticodon sequences? (2 points) Three (AGG, AGU, UCG). Codon–Anticodon Pairings Allowed by the Wobble Rules 5′ end of anticodon 3′ end of codon G C or U C G only A U only U A or G Problem set 2, BioS 220, 2011 Name _________________________ TA ______________________ 6 22. Give one example of each: (5 points total) a) Something cis-acting in prokaryotes? _operator or promoter_____________ b) Something cis-acting in eukaryotes? _promoter or enhancer or silencer______ c) Something trans-acting in prokaryotes? __ CAP or lacI repressor____ d) Something trans-acting in eukaryotes? __transcription factors___________ e) A type of modification that can happen to histone tails? __phosphorylation or acetylation or methylation or ubiquitination___________ 23. If a chromosomal inversion places the wild-type white+ allele INTO heterochromatin region, what do you expect to see in the mutant fly’s eyes? (1 point) a) The entire eye is red b) The entire eye is white c) Some eye facets are red and some are white d) Female flies will have red eyes and male flies will have white eyes 24. ___Chromatin remodeling_________________ is an ATP-dependent event that changes the interaction between DNA and nucleosomes, thereby changing the higher order structure of chromatin. (1 point) 25. Gene X encodes an enzyme that is expressed in the skin, the brain and the heart. Skinny, brainy, hearty are three genes that encode three tissue-specific transcription factors that activate the expression of gene X in the skin, the brain, and the heart, respectively. A student has isolated a mutant mouse in which the enzyme activity of X is greatly reduced in the brain, but is normal in the skin and the heart. The expression level of skinny, brainy, and hearty is normal in this mouse and there is no mutation in any of the three genes. Propose a simple explanation for the lack of X activity in the brain in the mutant mouse. (Assume that there is no alternative splicing for gene X in the three tissues.) (2 points) There may be a mutation in the enhancer that Brainy binds to. Since tissue-specific enhancers direct tissue-specific gene expression, this mutation will only affect the expression of X in the brain. Expression of X in the skin and the heart will not be affected. Problem set 2, BioS 220, 2011 Name _________________________ TA ______________________ 7 26. The first Lac promoter mutations that were isolated were “leaky” mutations that decreased, but did not entirely eliminate, the promoter function. The table below shows the behavior of such mutants giving the quantity of ß-galactosidase activity and Lac permease activity produced. ß-galactosidase activity permease activity lactose: – + – + Lac + (wild type) 2 100 2 100 LacP – 1 8 1 10 The P – mutations map very close to O c nevertheless researchers concluded that P and O probably represent functionally distinct sites on the DNA. Explain how the results shown above would lead to such a conclusion. (2 points) P- mutations result in decreased expression of Lac operon genes. Nevertheless, expression is still regulated by lactose. OC mutation, on the other hand, would result in unregulated expression of Lac operon genes. In other words, the activity of both beta-galactosidase and lac permease would be independent of lactose in OC. Because P is required for expression and the operator is involved in regulation of the operon, this implies that LacP and LacO represent functionally distinct sites on the DNA. (An alternative explanation – that P- is a mutation in LacO that results in a greater binding affinity with the Lac repressor – is not impossible, but unlikely. )