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CMMB 411 Lecture Notes - Lecture 7: Dna Replication, Thymidine Triphosphate, Hydrogen Bond

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rosefrog223
9 Oct 2018
School
University of Calgary
Department
Cellular, Molecular and Microbial Biology
Course
CMMB 411
Professor
Sui-Lam Wong

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Document Summary

The active site can fit the correctly base-paired nucleotides only. The active site has a particular shape, so it will select for the base pair that meets certain special requirements. There is a specific angle range (50 to 54 ) for correct base pairing, all other base pairs will have a different angle. If you have another combination (not watson and crick) of base pairs. Even if they have the optimal amount of hydrogen bonding, you cannot take (fit) those nucleotides because they will not fit into the pocket. Example: ac can have hydrogen bonding, but they won"t fit into the pocket because one or both requirements are not being met, so there is incorrect geometry. Dna polymerase sometimes will disincorporate the wrong nucleotide. However, according to the shape complementarity model, dna polymerase cannot make this replication error. The enzyme is dynamic, so it can change its shape depending on thermal energy.

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Related Questions

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.

1.A homozygous red-eyed female Drosophilamates with a red-eyed male Drosophila. What proportion of thefemale offspring will have white eyes?
75%
25%
100%
0%
2.Red-green colorblindness is a recessive X-linked trait.If a female is red-green colorblind, which of the following isTRUE?
Her father mustbe colorblind.
Both herparents are carriers of the recessive allele.
Both herparents must be colorblind.
Her mother mustbe colorblind.
Women cannotexhibit red-green colorblindness because they have two Xchromosomes.
3.Allelesare
recessive traits.
dominant traits.
exact copies of the same trait.
alternate versions of the same trait
4.Using three bases torepresent each amino acid in a polypeptide
ensures that other organisms can read the codetoo.
prevents the wrong amino acid from being incorporated into aprotein.
ensures that DNA can replicate properly.
ensures that all amino acids are represented at least once bythe code.
5. If an organism's DNA contains 20%adenine nucleotides, what percentage of its DNA is composed ofguanine nucleotides?
15%
10%
30%
40%
6. Atrue-breeding tall pea plant is crossed with a tall pea plant ofunknown parentage. Which of the following iscorrect?
Some shortindividuals may appear in the F2 generation.
Some of the F1generation will be short.
All of theoffspring (F1) will be short.
Over severalgenerations, no short individuals will appear.
7.Morgan breeds and sells snakes. He knows that he can makethe most money by breeding and selling albino snakes. Albinism is arecessive trait. His female albino corn snake just produced 6offspring. Two displayed normal coloration and 4 were albino. Whatdo you know about the genotype of the father?
He could beeither heterozygous or homozygous dominant.
He must behomozygous recessive.
There is notenough information to determine the father's genotype.
He must beheterozygous.
8.During the elongation cycle of translation, the A site ona ribosome functions in
releasing thecompleted polypeptide.
holding apolypeptide as amino acids are added.
attaching thesmall subunits of rRNA to the large subunit.
receiving a newtRNA with the correct amino acid.
9.Whichof the following is an incorrect match?
translation -polypeptide formation
transcription -tRNA formation
translation -mRNA binding to rRNA
replication -copying of DNA
10. Thethree steps that are required for DNA replicationare
base doubling,unwinding, and joining.
unwinding, basedoubling, and base pairing.
unwinding,complementary base pairing, and joining.
unwinding,transcription, and translation.
11.Thex-ray diffraction pattern of DNA discovered by Rosalind Franklinshowed that DNA
contains alphahelices and beta sheets.
contains betasheets arranged in a repeated pattern.
has a repeatedpattern.
is both helicaland has a repeated pattern.
12.Which of the following is TRUE?
All traits onthe sex chromosomes are associated with sexual development.
Females can beheterozygous for sex-linked traits.
Males areheterozygous for the SRY gene.
Men pass theirX-linked traits to their sons.
13.Martha has a widow's peak and attached earlobes. Martha's dad had astraight hairline and unattached earlobes. What is Martha'sgenotype?
Wwee
WwEE
wwee
WWEE
Answer withBRIEF EXPLANATIONS please
if more than 2answers you given are wrong, low star, pretty much no pts!