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HMM202- Final Exam Guide - Comprehensive Notes for the exam ( 26 pages long!)

699 views26 pages
indigohyena527
4 Dec 2017
School
Deakin University
Department
Health
Course
HMM202
Professor
Yann Gilbert

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Document Summary

Dna (cid:374)u(cid:272)leotides a(cid:396)e added i(cid:374) the (cid:1009)"-(cid:1007)" di(cid:396)e(cid:272)tio(cid:374). Dna (cid:396)epli(cid:272)atio(cid:374) is se(cid:373)i (cid:272)o(cid:374)se(cid:396)(cid:448)ati(cid:448)e. dna (cid:396)epli(cid:272)atio(cid:374) is diffe(cid:396)e(cid:374)t o(cid:374) the leading strand vs. the lagging strand (okazaki fragments). Single strand binding protein: maintain dna helix open: helicase: open dna double helix, rna primer: short fragment of nucleotide, dna polymerase iii: dna nucleotide adding enzyme which requires double stranded dna to be able to bind. Can only add nucleotide at the (cid:1007)"e(cid:374)d a(cid:374)d (cid:374)ot at the (cid:1009)"e(cid:374)d. the p(cid:396)o(cid:271)le(cid:373) with the other strand (the lagging) strand is i(cid:374) a (cid:1007)"-(cid:1009)" o(cid:396)ie(cid:374)tatio(cid:374). Dna at (cid:1007)"e(cid:374)d a(cid:374)d fill the (cid:271)la(cid:374)ks (cid:894)okazaki fragents). Dna polymerase then fills the gap a(cid:374)d uses dna ligase to (cid:862)glue(cid:863) the dna pieces together. 3% of dna encode protein coding gene, a lot of non-coding dna. Only protein coding genes are transcribed in rna (mrna). Eukaryote genes are made of introns and exons but only exons are translated into proteins. In eukaryotes: 1 gene = 1mrna = 1 protein.

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Related Questions

Multiple choice questions:

1) For DNA Polymerase III to work, which of these components has to act first? (A. Conservative replication, B. Newly replicated strand, C.Primase, D. RNA, E.Mutation rate)

2) The double helix of DNA is wrapped about 1 3/4 times around which of these components? (A. Helicase, B. DNA polymerase III, C. DNA polymerase I, D.Nucleosome, E. DNA ligase)

3) Which strand is produced more rapidly? (A.Okazaki fragment, B. Lagging strand, C. Leading strand, D. 3' , E. 5' )

4) During DNA replication in the lagging strand there is production of this component? (A. Okazaki fragment, B. mRNA, C. tRNA, D. proteins, E. none of the above)

5) Which enzyme is used first to permit DNA polymerase III to begin replicating the 3' stand? (A. Conservative replication, B. Parental stand, C. Primase, D. RNA, E. Mutation rate)

6) Which enzyme unwraps the double helix? (A. Helicase, B. DNA polymerase III, C.DNA polymerase I, D. Nuclesome, E. DNA ligase)

Thanks for any feedback, want to verify my answers.

Double check to see if I got these right? 3 questions

Some Processes Involved in DNA Replication

1.The replication origin is identified.

2 DNA primase builds RNA primer.

3Okazaki fragments are spliced by DNA ligase.

4Helicases bind and uncoil the DNA double helix.

5 RNA primer is the starting point for DNA polymerase.

6 DNA replication ends at the telomere where DNA codes fortermination.

7 DNA polymerase adds complementary nucleotides in the 5' to 3'direction in short segments called Okazaki fragments.

The sequence in which the events numbered above occur during DNAreplication of the lagging strand: Number 1 - 7 in order.

I put: 1,4,2,5,7,3,6

__________________Question 2:

Some Functions of Enzymes

  1. Build DNA strands
  2. Build RNA primers
  3. Build mRNA strand
  4. Cleave DNA strands
  5. Splice Okazaki fragments

Enzyme:
Helicase :4
DNA ligase :5
RNA primase :1
RNA polymerase :3
DNA polymerase:2

Question 3:

like mRNA, tRNA has a ribose sugar, U instead of T, and issingle stranded. Unlike mRNA, which remains a long single strand ofnucleotides, tRNA folds so that some areas pair up.

The resulting structure has an anticodon on one end and a site foran amino acid to attach on the other end. There is basecomplementarity (A pairs with U and G pairs with C) between an mRNAcodon and tRNA anticodon.

If the amino acid tyrosine attaches to a tRNA, which of thefollowing anticodons could be at the opposite end of the tRNAmolecule?

Select one:

a. ATA and ATG

b. UAU and UAC

c. AUA and AUG

d. AGA and AGU

I picked B.

1. In addition to identifying the genetic material, the experiments of Avery, MacLeod, and McCarty with different strains of Streptococcus pneumoniae demonstrated that
DNA is a double helix.
DNA may be taken up by bacterial cells and be active.
DNA is present in bacteriophage T2.
eukaryotic cells may be genetically transformed.
DNA binds to the cell wall of the bacterium.
2. In order to show that DNA in cell extracts is responsible for genetic transformation in Streptococcus pneumoniae, important corroborating evidence should indicate that _______ also destroy transforming activity.
temperatures high enough to kill cells
enzymes that hydrolyze proteins
enzymes that hydrolyze DNA
enzymes that hydrolyze polysaccharides
enzymes that hydrolyze RNA
3. Based on what you have learned about the experiments conducted by Griffith and Avery and colleagues with bacteria, which of the following would result in transformation of living R cells?
Pure RNA from S cells
Cell-membrane extracts from S cells
A mixture of pure RNA and protein from S cells
Cell-free extracts from S cells
Pure protein from S cells
4. A-T base pairs in a DNA double helix
form three hydrogen bonds with each other.
are more heat-stable than G-C base pairs.
are of a substantially different length than G-C base pairs.
are not accessible to DNA binding proteins.
are chemically distinct from G-C base pairs.
5. If 23 percent of the bases in a sample of double-stranded DNA are adenine, what percentage of the bases are uracil?
27
50
0
73
23
6. The uniform diameter of the DNA structure provides evidence for
antiparallel DNA strands.
a right-handed helix.
a double-stranded helix.
3′ end phosphorylation.
purine–pyrimidine pairing.
7. If a sequence of one strand of DNA is 5′-TGACTATC-3′, what is the complementary strand?
5′-ACTGATAG-3′
3′-TGACTATC-5′
3′-ACTGATAG-5′
3′-ACTGATAG-3′
5′-TGACTATC-3′
8. What structural aspect of the DNA facilitates dissociation of the two DNA strands for replication?
3′ end phosphorylation
Purine–pyrimidine pairing
Antiparallel DNA strands
Covalent bonds between sugars and phosphate groups
Hydrogen bonding between paired bases
9. If the Meselson–Stahl density gradient experiment had resulted in two bands of DNA molecules after only one round of replication, one containing only 15N and the second only 14N, this result would have indicated that replication was
unidirectional.
dispersive.
conservative.
bidirectional.

semiconservative.
10. The nucleoside analogue acyclovir, which is used to treat herpes simplex virus (HSV) infections, lacks a 3′ hydroxyl group (–OH). Predict what will happen if the host cell DNA polymerase incorporates a molecule of acyclovir into an elongating strand of HSV DNA.
The replication complex will no longer bind to origins of replication.
The DNA polymerase will only incorporate acyclovir molecules.
Acyclovir will bind single-strand binding proteins.
DNA polymerase will not be able to link a successive nucleotide.
Acyclovir will block the activity of the DNA helicase.
11. Which of the following does not demonstrate the stability of the DNA double helix?
Single-strand binding proteins are required to keep the strands apart.
Paired bases form hydrogen bonds.
High temperatures are required to denature it.
DNA synthesis requires energy.
DNA helicases require ATP.
12. What effect would a primase inhibitor have on DNA replication?
DNA polymerase would not be able to add bases to the DNA.
Primase would be blocked from joining the DNA replication complex.
Primase would not be able to provide primers for DNA polymerases.
DNA polymerase would not be able to use DNA as a template.
There would be no effect on DNA replication.
13. To replicate their DNA in a timely manner, most eukaryotic chromosomes
normally have uncoiled DNA.
have multiple origins of replication.
contain linear molecules of single-stranded DNA.
have two replication forks that move in the same direction.
are composed entirely of DNA.
14. Which statement about DNA replication is false?
Okazaki fragments are synthesized as parts of the leading strand.
Error rates for DNA replication are reduced by proofreading of the DNA polymerase.
The sliding clamp increases the rate of DNA synthesis.
Replication forks represent areas of active DNA synthesis on the chromosomes.
Ligases and polymerases function in the vicinity of replication forks.
15. In many eukaryotes, there are repetitive sequences called telomeres at the ends of chromosomes. After successive rounds of DNA replication, the _______ strand becomes shorter. In some cells, an enzyme called _______ repairs the shortened strand.
leading; telomerase
lagging; telomerase
lagging; DNA polymerase I
leading; DNA polymerase I
leading; replicase
16. A researcher studies normal human fibroblast cells. They can be maintained in culture but die off after about 30 cell generations. Unexpectedly, a colony of cells continues to survive and divide past 30 generations. Which scenario is most likely true for these cells?
DNA polymerase is constantly active.
Primase is able to extend the telomeres.
The mismatch repair system is extra efficient.
The telomerase gene is being expressed.
The cells do not need the ends of their chromosomes.
17. If DNA polymerase III introduces an incorrect nucleotide, what is the first corrective action made by the DNA repair system?
The replication complex excises the incorrect nucleotide.
The excision repair proteins remove the incorrect nucleotide.
The mismatch repair system scans for base-pairing mismatches.
There is no correction, because nucleotide errors by DNA polymerases are not repaired.
DNA ligase connects the correct nucleotide.
18. Choose the correct order of the following four events in the excision repair of DNA:
(1) Base-paired DNA is made complementary to the template.
(2) Damaged bases are recognized.
(3) DNA ligase seals the new strand to existing DNA.
(4) Part of a single strand is excised.
1, 2, 3, 4
2, 1, 3, 4
3, 4, 2, 1
2, 4, 1, 3
4, 2, 3, 1
19. Six complete cycles of PCR should result in a _______-fold increase in the amount of DNA.
64
32
12
128
6
20. When double-stranded DNA is heated to temperatures above 90°C, it denatures. Denaturation is a process that
breaks covalent bonds.
hydrolyzes DNA.
separates double-stranded DNA into two single strands.
causes new hydrogen bonds to form.
irreversibly destroys the double helical structure.