HMB265H1 Lecture Notes - Lecture 17: Mutation Rate, Dna Replication, Pyrimidine
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QUESTION 1
A mutation caused by exposure to gamma rays is called a spontaneous mutation
a. | True | |
b. | False | |
c. | Sometimes, depending on the type of base change | |
d. | Gamma rays do not cause mutations |
QUESTION 2
A recessive mutation affecting an essential biochemical pathway can be detected________
a. | when the organism is heterozygous for that mutation. | |
b. | when the organism is homozygous for that mutation. | |
c. | in either homozygous or heterozygous conditions. | |
d. | only by sequencing (recessive mutations never show a phenotype). |
QUESTION 3
During DNA replication, the newly synthesized chain grows by adding
a. | nitrogenous bases | |
b. | the sugar-phosphate backbone | |
c. | nucleosides | |
d. | dNTPs |
QUESTION 4
During replication, the DNA strand 5â- AAGTCTAGCCTAG -3â will serve as a template for the polymerization of:
a. | 5â- CTAGGCTAGACTT -3â | |
b. | 5â - TTCAGATCGGATC -3â | |
c. | 5â - GATCCGATCTGAA -3â | |
d. | 3â - AAGTCTAGCCTAG -5â |
QUESTIOn 5
Given the following DNA sequences, select which statement(s) is(are) correct.
Sequence 1:
5â-TGGACGCTAA-3â
3â-ACCTGCGATT-5â
Sequence 2:
5â-AATCGCAGGT-3â
3â-TTAGCGTCCA-5â
Sequence 3:
5â-ACCTGCGATT-3â
3â-TGGACGCTAA-5â
a. | Sequences 1 and 2 are the same | |
b. | Sequences 1 and 3 are the same | |
c. | Sequences 2 and 3 are the same | |
d. | Sequences 1, 2 and 3 are all different |
QUESTION 6
Repetitive sequences in the genome are hotspots for:
a. | Deamination | |
b. | Depurination | |
c. | Thymine dimer formation | |
d. | Replication errors |
QUESTION 7
Select which statement(s) about the Ames test is(are) correct
a. | it has been designed to understand mutation repair systems in Salmonella typhimurium | |
b. | it uses mammalian liver extract | |
c. | it allows to study whether chemical compounds or their enzymatic breakdown products are mutagenic | |
d. | it is based on whether a chemical compound causes reversion mutations from his+ to his- | |
e. | two of the above are correct | |
f. | three of the above are correct |
QUESTION 8
Select which statement(s) is(are) correct.
a. | All DNA strands have a direction, and it is specified by the carbons in the sugar backbone. | |
b. | All DNA strands have a direction, and it is specified by hydrogen bonds between nucleotides. | |
c. | In order to form proper base pairs in a double stranded DNA molecule, the two strands must run in opposite directions. | |
d. | A and C are correct. | |
e. | B and C are correct. |
QUESTION 9
Studies of gene mutation frequencies have shown that:
a. | mutations are rare, and genomes are generally stable. | |
b. | mutation frequencies differ among organisms and also between genes, suggesting certain genes are more susceptible to mutation. | |
c. | mutation frequencies are consistent between organisms, and each region of DNA is equally susceptible to random mutations. | |
d. | Both A and B are correct. | |
e. | Both A and C are correct. |
QUESTION 10
The compound 5-bromodeoxyuridine (BrdU) is a derivative of uracil, and if BrdU becomes incorporated during DNA replication, it pairs with adenine. This compound is best classified as which type of mutagen?
a. | base analog | |
b. | base inducer | |
c. | intercalating agent | |
d. | oxidative agent | |
e. | alkylating agent |
QUESTION 11
The rate of mutation of the fruit fly is higher than the rate of mutation of Algae
True
False
QUESTION 12
Thymine dimers are most commonly caused by which of the following?
a. | X-rays | |
b. | Alkylating agents | |
c. | U.V. irradiation | |
d. | DNA intercalating agents |
QUESTION 13
What chemical group is found in the 3â end of a DNA strand?
a. | an alcohol | |
b. | a hydroxyl | |
c. | a methyl | |
d. | a phosphate |
QUESTION 14
What chemical group is found in the 5â end of a DNA strand?
a. | an alcohol | |
b. | a hydroxyl | |
c. | a methyl | |
d. | a phosphate |
QUESTION 15
Which is the correct order of molecules binding to DNA during DNA replication?
a. | Helicase, SSB, primase, DNA pol III, DNA pol I, ligase | |
b. | SSB, DNA pol I, ligase, helicase, DNA pol I, primase | |
c. | primase, helicase, DNA pol III, ligase, DNA pol I, SSB | |
d. | SSB, helicase, primase, DNA pol III, DNA pol I, ligase |
QUESTION 16
Which statement(s) is(are) correct about strand slippage?
a. | It can cause disorders such as Huntingon disease. | |
b. | It is a process that causes mutations altering the number of DNA repeats. | |
c. | It is a process that incorporates nucleotide base analogs and trinucleotide repeats. | |
d. | a and b are correct. | |
e. | b and c are correct. | |
f. | a, b and c are correct. |
help needed asap.
Homozygous recessive loss of function mutations in any one of 4 different genes in C. elegans worms (genes A, B, C, or D) results in female worms that don't form proper egg-laying structures, called vulvas, as shown in Lines 2-5 of the data table below. In vulvaless females, the fertilized eggs hatch and baby worms develop inside the mother, killing her in the process.
genotype | phenotype | |
1 | AA BB CC DD (wild type) | normal vulva |
2 | aa BB CC DD | No vulva |
3 | AA bb CC DD | no vulva |
4 | AA BB cc DD | no vulva |
5 | AA BB CC dd | no vulva |
You would like to figure out the order in which these genes act during the creation of the vulva (you cannot assume it will be A--->B--->C--->D as the gene names are arbitrary). Your friend tells you to perform epistasis analysis by making double mutants between the different homozygous recessive mutants and analyzing the phenotype of the double mutant. For example, she asks you to examine the phenotype of aabbCCDD (double mutant of genes A and B) and compare this to the single mutants to figure out whether gene A acts earlier than gene B, or vice versa. You know this won't work. Why not? Pick the ONE BEST choice:
a. The phenotype of the single mutant is already pretty severe. The double mutant will most likely be dead, therefore epistasis analysis won't be possible.
b. Each of the single mutants (aaBBCCDD) and (AAbbCCDD) has the same mutant phenotype i.e., no vulva. Epistasis analysis between any 2 genes is only possible when the mutant phenotypes for each gene is different.
c. Epistasis analysis involves making triple mutants (such as aabbccDD) in order to learn something about how the genes are ordered.
d. Epistasis analysis can never be carried out with null loss of function mutations. The mutations being analyzed all have to be dominant gain of function alleles.
To address your concern, you first generate overactive alleles of either gene A (denoted as A*) or gene B (denoted as B*). As shown in lines 6-7 of the table below, C. elegans that have one of these overactive alleles produce multiple vulvas.
Genotype | Phenotype | |
6 | A*A BB CC DD | multiple vulvas |
7 | AA B*B CC DD | multiple vulvas |
To find out the order in which these genes act, you combine the overactive alleles with different loss of function alleles and observe the phenotype in double mutants (see Lines 8-11).
Genotype | Phenotype | |
8 | A*A bb CC DD | multiple vulvas |
9 | A*A BB cc DD | multiple vulvas |
10 | A*A BB CC dd | no vulva |
11 | AA B*B cc DD | multiple vulvas |
Based on the data in the table, what is the order in which these 4 genes normally act in wild-type C. elegans in order to produce a wild-type/normal vulva?
a. A----> B----->C----->D ---> Vulva
b. D----> B----->C----->A----> Vulva
c. B----> C----->A----->D----> Vulva
d. C----> B----->A----->D----> Vulva
e. C----> B----->D----->A----> Vulva
f. don't have enough data to make any conclusions
Based on the vulva formation phenotype of the double mutants, which of the following statement(s) accurately describes the genetic interactions between the A* allele and alleles of other genes affecting vulva formation? Pick ALL that apply:
a. The A* allele is epistatic to homozygous recessive loss of function mutations in gene B
b. A homozygous recessive loss of function mutation in gene B is epistatic to the A* allele
c. Homozygous recessive loss of function mutation in gene D is epistatic to the A* allele
d. The A* allele is epistatic to homozygous recessive loss of function mutation in gene D
e. The A* allele enhances the homozygous recessive loss of function mutation in gene D