Mock Exam 2

10/18/15

  1. Which of the following is NOT typical of prokaryotes?
  1. Their DNA is consisted of one chromosome
  2. Their DNA has one origin of replication
  3. They have circular a chromosome called a nucleoid
  4. They have multiple origins of replications

Pages 301, 329

  1. True/False- mRNA reads DNA downstream from the transcription start site.

Page 362 and 371. The transcription start site is the first nucleotide that is read by RNA polymerase seen as +1. From here, mRNA goes downstream and transcribes the gene on the template strand of DNA. The consensus sequences will be upstream about -25 nucleotides for eukaryotes and -10 and -35 for prokaryotes.

  1. Which of the following is true for germ-line mutation?
  1. They occur in non-reproductive cells
  2. They occur in about half of the members of the next generation
  3. Only a few cells will carry the mutation
  4. All of the above

Since germ-line mutation occurs in gametes, in the next generation an individual will either have all of the cells with the mutation or none of the cells will have the mutation depending on meiosis. Page 494-495

  1. True/False- Transversions are easier mutations to occur than transitions, because the number of possible transversions is twice the number of possible transitions.

False- Transforming a purine into a different purine (transition) is easier to transform rather than transforming a purine into a pyrimidine (or pyrimidine to purine) page 495

  1. Suppose there is a base substitution in the coding strand of DNA from TCA to TAA. This will be seen as what kind of mutation?
  1. Missense
  2. Nonsense
  3. Silent
  4. Frameshift

Page 498 and 420. To answer this problem, we need to know what happens to the amino acid that will be translated. TAA on the coding strand leads to an mRNA with the sequence UAA. This is one of the three stop codons in translation, the other two being UGA and UAG. Since this mutation stops translation prematurely, it is a nonsense mutation.

  1. Which of the following is an example of posttranscriptional modifications of eukaryotic pre-mRNA?
  2. Addition of a 5’ cap
  3. Termination of transcription via the Rho protein
  4. Synthesis of a short primer via RNA primase
  5. All of the above

Page 389- note that post-transcriptional modifications happen after transcription, hence the name. Also, use of the Rho protein is seen only in prokaryotes and RNA primase is seen in DNA synthesis.

  1. If a double-stranded DNA molecule is 15% guanine, what is the percentage of thymine?
  2. 35%
  3. 15%
  4. 30%
  5. 70%

We know that A=T and G=C. We also know that A + C = G + T. If we have a molecule with 15% guanine, we know that there will be the same amount of cytosine because G=C. So there will be 15% cytosine. This means that G + C = 30%. If that’s the case, the remaining 70% is split between A and T so thymine is 35%.

  1. If you have a gene within some heterochromatin, why is this not optimal?
  2. Heterochromatin is not normally transcribed so that gene will most likely not be read
  3. Heterochromatin has many genes and the RNA polymerase will become confused on where to start transcription
  4. Heterochromatin is less condensed, so it is harder for the gene to be transcribed
  5. Heterochromatin is located along chromosome arms which isn’t optimal

Page 302, know table 11.1

  1. Which of the following is true about telomeres?
  2. They protect the sticky ends of chromosomes
  3. They provide a means of replicating the ends of chromosome
  4. Telomeres are in the chromosomes of protozoans, humans and plants
  5. All of the above

Page 307

  1. Which of the following proteins relieves strain from the ahead of the replication?
  2. DNA helicase
  3. DNA gyrase
  4. Single-stranded-binding protein
  5. RNA primase

Page 335

  1. What is the function of DNA ligase?
  2. Joins Okazaki fragments
  3. Prevent secondary structures from forming on single-stranded DNA
  4. Binds to origin and separates strands of DNA to initiate replication
  5. Removes RNA primers and replaces them with DNA

Page 338 know table 12.4

  1. If a double-stranded RNA molecule had 20% adenine, what is the percentage of thymine?
  2. 20%
  3. 40%
  4. 10%
  5. 30%
  6. None of the above

Uracil replaces thymine in RNA so there will be 20% uracil and 0% thymine. There will be a combined 60% between cytosine and guanine where each will be 30% in the molecule. A=U in RNA.

  1. What would happen if primase was not found in a bacterial cell?
  2. DNA replication wouldn’t occur
  3. DNA wouldn’t be unwound at the replication fork
  4. DNA wouldn’t separate
  5. Okazaki fragments wouldn’t be joined together

Page 338

  1. Are introns part of a gene?
  2. No because introns don’t code for amino acids
  3. Yes because the introns are part of the original nucleotide sequence
  4. Yes because introns code for amino acids
  5. No only exons are in pre-mRNA

Page 387 only exons are used in mature mRNA to be translated into amino acids

  1. True/False: The 5’ and 3’ untranslated regions are part of the exons in pre-

mRNA.

On a gene, the first and last exon are part of the untranslated region and the rest of the introns and exons will be in between these two exons. Page 385-386

  1. All of the following are in prokaryotes and eukaryotes except:
  1. DNA polymerase
  2. Helicase
  3. RNA polymerase
  4. Telomerase
  5. Primase

Prokaryotes have circular chromosomes, there are no telomeres

  1. On what kind of RNA is a methyl-G cap found?
  2. mRNA
  3. rRNA
  4. tRNA
  5. A and B are correct
  6. All of the above are correct

Page 389 mRNA goes through post transcriptional modifications, tRNA is used in translation, giving the correct amino acid to the rRNA which reads mRNA

  1. Given a strand of DNA, choose its complimentary strand.

5’ ACGTGC 3’

  1. 5’ TGCACG 3’
  2. 5’ GCACGT 3’
  3. 5’ GCGACG 3’
  4. 5’ TGCTCG 3’
  1. To which carbon is the base attached in a nucleotide?
  2. 5’
  3. 3’
  4. 2’
  5. 1’

Page 286

  1. Histone proteins can undergo various modifications such as ______which ______
  1. Acetylation – increases transcription
  2. Methylation – decreases transcription
  3. Carboxylation – increase transcription
  4. A and B
  5. A, B, and C
  1. True/False: Frameshift mutations always alter every amino acid following the mutation.

False: some frameshift mutations consist of multiples of three. This may still affect the phenotype, however the frame stays intact so all codons following the frameshift will stay the same page 496

  1. True/False: Intragenic suppressor mutations fix frameshift mutations by inserting another nucleotide somewhere along a sequence that had a deletion.

True: It may not add to the same spot as the deletion, so there may be a couple of amino acids affected, but the rest of the frame afterwards is restored to its original form page 499

  1. Which of the following is true about transposable elements?
  1. They are sequences that can move about the genome
  2. They make up about 45% of bacterial DNA
  3. They are found in the genomes of all organisms
  4. Only two above are correct
  5. All of the above

Page 511 b is incorrect, that is supposed to be human DNA. Although they are in bacterial DNA, I didn’t see a general percentage that all bacteria share so correct me if I’m wrong. A and C are true about transposable elements.

  1. Sigma factor is to RNA polymerase in prokaryotes as ______is to ______in eukaryotes
  1. RNA polymerase I; initiation
  2. TATA-binding protein; RNA polymerase
  3. Core promoter; mRNA
  4. Rho-independent; termination

Page 371 and 372. So when a sigma factor binds to an RNA polymerase it initiates transcription at the proper site in prokaryotes, this is typically along the promoter region (page 364), for eukaryotes, however, the TATA-binding protein will find the TATA box in eukaryotes which bends the DNA and partially unwinds it allowing more transcription factors to assemble on the promoter as well as the RNA polymerase

  1. Draw out the leading strand and lagging strand on a DNA molecule that is undergoing replication, label the 5’ and 3’ ends for all four strands.

Page 334 figure 12.10

  1. Draw out the template and non-template strands of DNA from the mRNA strand that reads – 5’-AAUGCGGCUAGCGUAUGCCGAUGCA-3’

Template 3’-TTACGCCGATCGCATACGGCTACGT-5’

Non template 5’-AATGCGGCTAGCGTATGCCGATGCA-3’

  1. List at least three mutations and give an example or definition for each.
  1. Transitions- Base change from A to G
  2. Transversions- base change of one pyrimidine to a purine (or vice versa) T to A
  3. Missense mutation – change from one amino acid to the next Ex. TCA to TTA which changes the codon
  4. Nonsense mutation- change in a codon that makes it a stop codon Ex. TCA to TAA which codes for UUA which is a stop codon in translation
  5. Silent mutation- a base change that codes for the same amino acid
  6. Frameshift mutation- insertion or deletion in the reading frame that changes each amino acid and can cause translation to stop prematurely
  1. Describe the basic function of mRNA, tRNA and rRNA in translation.

You know how it goes DNA  RNA  protein (product). Translation is the arrow between RNA and product. We use translation by adding amino acids together one at a time to make a polypeptide chain.

tRNA- deliver amino acids to ribosomes, it carries amino acids to the rRNA and mRNA complex and adds amino acids to the growing chain. It has what is called an anticodon that reads the mRNA and gives the growing chain the correct amino acid by reading the codon on the mRNA

mRNA- carries coding instructions for a polypeptide chain made which will be made up of amino acids

rRNA- ribosomal RNA moves down the mRNA and “reads” it. It reads the mRNA codon by codon and has each tRNA with the correct anti codon sequence add its amino acid to the growing polypeptide chain.

Pages 426-429, 424

  1. Why are multiple primers needed on the lagging strand in DNA synthesis?

DNA polymerase elongates from the 3’-OH group provided by a short RNA primer. Also, DNA synthesis runs in the 5’3’ direction. The leading strand is able to keep synthesis going continuously, because it is going in the same direction as that of unwinding. The lagging strand is because after a short length of DNA is unwound, synthesis must proceed in the 5’3’ direction, but here it is opposite that of unwinding of DNA. Only a short section of DNA is needed for synthesis to start, so as it keeps unwinding, more and more primers are being put on so that DNA can replicate in the 5’3’ direction. Page 333-334

  1. List as many components as you can that are required for DNA replication and give their function.
  1. Initiator protein- separates strands of DNA to initiate replication
  2. DNA helicasae- unwinds DNA
  3. Single-stranded-binding proteins- keeps secondary structures from forming
  4. DNA gyrase – releases torque that builds up as a result of unwinding
  5. DNA primase- synthesizes a short RNA primer to provide a 3’-OH group for the attachment of nucleotides
  6. DNA polymerase- elongates new nucleotide strand, removes RNA primers and replaces them with DNA
  7. DNA ligase- joins okazaki fragments (puts together the lagging strand)