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MBG 2040 Chapter Notes - Chapter 10: G1 Phase, Central Dogma Of Molecular Biology, Lambda Phage

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rubylocust666
2 Dec 2016
School
University of Guelph
Department
Molecular Biology and Genetics
Course
MBG 2040
Professor
Baozhong Meng

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Document Summary

Genes are found in the form of dna. Rna is translated into polypeptides (amino acids) The central dogma was discovered by francis crick in 1958. Reverse transcriptase found in retro viruses and retro-like viruses defies the central dogma. Consists of: anti-parallel strands, complementary base pairs. Bonded with only two hydrogen bonds making these bonds weaker. Other possible methods of replication include: conservative: In conservative replication, the parental double helix would be conserved, and a new progeny double helix would be synthesized: dispersive: In dispersive replication, segments of both strands of the parental dna molecules would be conserved and used for the synthesis of complimentary segments that would join together to produce the progeny dna. Meselson and stahl proved that semi-conservative model was the correct mechanism for replicating dna. 6m cscl = 1. 7 g/cm3 ; 14n e. coli dna = 1. 710 g/cm3 ; 15n e. coli dna = 1. 724 g/cm3.

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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.

13. The arms of a human and the wing of a bat are _________________ structures, composed of the same bones that have been modified for different functions.

a. analogous
b. vestigial
c. homologous
d. convergent

14. Xeroderma pigmentosum is an autosomal recessive genetic disorder. People who are homozygous for the disease-causing allele have skin that is damaged by exposure to ultraviolet radiation; this can lead to skin cancer at an early age. The mutation that causes this disorder is in a gene that functions:

a. to repair DNA by excising (removing) nucleotides damaged by UV light.
b. in lysosomes, to break down lipids that otherwise accumulate in the nervous system.
c. to produce the UV blocking pigment melanin in the skin cells
d. in red blood cells to carry oxygen to the skin.

15. The Amish are a religious sect descended from a small group of colonists who came to the United States over 200 years ago. Their religious beliefs have kept them isolated from society. The Amish have a much higher than average incidence of both polydactyly (extra fingers and toes) and dwarfism in their communities. This is most likely explained by:

a. inheritance of acquired characters.
b. the founder effect.
c. disruptive selection.
d. natural selection.

16. In the structure of DNA that Watson and Crick proposed, the sides of the "ladder" of the DNA molecule are always the same distance apart, making the molecule a constant width. This occurs because when the bases hydrogen bond to form the rungs of the "ladder", a:

a. purine always pairs with a pyrimidine.
b. purine always pairs with a purine.
c. pyrimidine always pairs with a pyrmidine.
d. The structure of the double helix is not determined by how the bases pair.

17. The RNA transcript of DNA that travels to the cytoplasm, carrying the instructions to make a protein, is called:

a. rRNA.
b. tRNA.
c. mRNA.
d. RNA polymerase.

18. When DNA is replicated, the error rate is approximately one error for every 10,000 nucleotides copied. However, that error rate is reduced to only 1 error for every 1 billion nucleotides. The DNA sequence is "corrected" by:

a. repeating S phase to see if the error rate is lower the second time around.
b. enzymes that proofread the DNA and repair errors.
c. messenger RNA during the process of transcription.
d. Any errors made during DNA replication cannot be corrected. If errors are detected after S phase, the cell is destroyed.

19. Mendel's principle of segregation says that:

a. when gametes are formed, each gamete receives only one allele for a particular gene.
b. some genes are dominant to others.
c. a testcross must be used to determine the genotype of an organism with a dominant phenotype.
d. all of the above

20. Tay-Sachs disease is lethal before reproductive age, but the allele persists in Louisiana French Canadians because:

a. it is a dominant allele.
b. new mutations causing this disease are common in that population.
c. it is not expressed in the phenotype of heterozygous individuals.
d. the disease is contagious and can be transmitted by nonheritable means.

21. Physical features that are similar in organisms which are not closely related to each other are most likely a result of ____________. A good example of this is the similarities seen between some marsupial and placental mammals.

a. natural selection.
b. adaptation to similar environments
c. convergent evolution
d. All of the above phrases can be used to correctly fill this blank.

22. In humans, which of the following sex chromosome compliments could be found in males?

a. XY
b. XX
c. Y

23. The proteins that are associated with eukaryotic chromosomes:

a. help to package the DNA within the nucleus.
b. are involved in gene expression, through binding with the DNA.
c. may be replicating or transcribing the DNA.
d. All of these are true of proteins that are associated with eukaryotic chromosomes.

24. A sequence of DNA nucleotides that contains the information to produce a single protein is a(n):

a. codon.
b. gene.
c. polypeptide.
d. anticodon.

1. Characters that show a continuous range of variation, such as height and eye color, usually are controlled:

a. by a single gene with two alleles that are codominant.
b. by many genes with an additive effect.
c. by epistatic interactions between two genes.
d. mainly by the environment, with only a small genetic component.

2. In humans, red-green colorblindness is inherited as a sex-linked recessive trait. In order for a woman to be red-green colorblind, which of the following statements must be true.

a. Her mother must be red-green colorblind.
b. All of her brothers must be red-green colorblind.
c. Her father must be red-green colorblind.
d. All of the above statements must be true if a woman is red-green colorblind.

3. The x-ray crystallography data collected by Rosalind Franklin suggested to Watson and Crick that the:

a. structure of DNA is a double helix.
b. two strands of the DNA molecule are joined by hydrogen bonds between the bases.
c. four bases within DNA pair in a specific way.
d. two strands of the DNA molecule are joined by covalent bonds between the bases.

4. In the genetic code, _________ one amino acid.

a. one nucleotide specifies
b. two nucleotides specify
c. three nucleotides specify
d. four nucleotides specify

5. During Meiosis I, a homologous pair of chromosomes may not separate, resulting in daughter cells that have extra chromosomes or are missing chromosomes. This can lead to genetic disorders, including Down Syndrome. This phenomenon is called:

a. independent assortment.
b. nondisjunction.
c. segregation.
d. crossing over.

6. You are a human geneticist studying the incidence of retinitis pigmentosa in the residents of Tristan de Cunha, a group of small islands in the middle of the southern Atlantic Ocean. The allele for retinitis pigmentosa, which causes a form of blindness, is inherited as an autosomal recessive. You have determined that the frequency of this allele (r) in the population is 0.4 (40%). Using the principles of the Hardy-Weinberg rule, you would estimate the frequency of individuals who are heterozygous for this allele (Rr) in the population to be:

a. 0.16 (16%)
b. 0.24 (24%)
c. 0.36 (36%)
d. 0.48 (48%)

7. Natural selection acts at the level of the:

a. phenotype.
b. gene.
c. population.
d. nucleotide.

8. You are working with pea plants, trying to recreate the experiments that Mendel performed. You are doing a dihybrid cross with a plant that is heterozygous for both seed shape and seed color, with the genotype RrYy. Which allelic combinations would you expect to find in the gametes produced by this plant?

a. This plant would produce only RY and ry gametes.
b. This plant would produce only RrYy gametes.
c. This plant would produce RY, Ry, rY, and ry gametes.
d. You cannot determine which gametes this plant can produce without knowing the genotypes of its parents.

9. Biochemist Erwin Chargaff found that in DNA there is a special relationship between the four bases that we now call Chargaff's rule. His observation was that, in an organism's genome the:

a. percentage of A nucleotides = the percentage of T nucleotides, and the percentage of C nucleotides = the percentage of G nucleotides.
b. four bases all occur in an equal frequency (25%) within each organism.
c. percentage of A nucleotides = the percentage of G nucleotides, and the percentage of C nucleotides = the percentage of T nucleotides.
d. genetic material is composed of proteins, not DNA.

10. During DNA replication:

a. each strand of the double helix acts as a template for the synthesis of a new strand.
b. the enzyme DNA polymerase adds nucleotides to the strand being synthesized.
c. the bases A,C,G and T are required.
d. All of the above are true of DNA replication.

11. During translation, amino acids are joined by peptide bonds to make polypeptides. The formation of these peptide bonds is catalyzed by:

a. DNA.
b. mRNA.
c. tRNA.
d. rRNA.

12. If an allele (R) at a gene with two alleles shows complete dominance, individuals with the genotypes ______ will have the same phenotype.

a. RR and rr.
b. RR and Rr
c. Rr and rr
d. Each of the three possible genotypes will have a different phenotype.