Question 1

Experiment 1 depends on recombination.

Accepted Answer

The answer of Experiment 1 depends on recombination....

Question 2

Fruit flies normally have red eyes.Seven different true-breeding strains of fly with white eyes have been identified (A-H).In each strain, the white eye trait is due to an autosomal recessive allele.It is possible that all seven strains have mutations in the same gene.Alternatively, some or all of the strains may have mutations in different genes.To determine how many genes are involved in eye color in these flies, all possible pairwise crosses are performed.The offspring phenotypes resulting from each cross are observed.(+ = wild-type; ? = mutant)
 $$\begin{array} { | c | c | c | c | c | c | c | c | c | } 
\hline & \mathrm { A } & \mathrm { B } & \mathrm { C } & \mathrm { D } & \mathrm { E } & \mathrm { F } & \mathrm { G } & \mathrm { H } \
\hline \mathrm { A } & - & - & + & + & - & + & + & - \
\hline \mathrm { B } & & - & + & + & - & + & + & - \
\hline \mathrm { C } & & & - & - & + & + & + & + \
\hline \mathrm { D } & & & & - & + & + & + & + \
\hline \mathrm { E } & & & & & - & + & + & - \
\hline \mathrm { F } & & & & & & - & - & + \
\hline \mathrm { G } & & & & & & & - & + \
\hline \mathrm { H } & & & & & & & & - \
\hline
\end{array}$$
-Which strains' mutations represent a complementation group?
A)C, D, F
B)A, B
C)E
D)A, B, E, G, H

Accepted Answer

A complementation group is formed by mutations that do not complement each other, meaning they are likely in the same gene. In this table, strains A, B, E, G, and H do not complement each other (indicated by "-" when crossed), suggesting they are mutations in the same gene.

Question 3

A bacterial mismatch repair system is able to correct replication errors that insert an incorrect nucleotide.How is this repair system able to determine which mismatched base is incorrect?
A)The incorrect base results in a bulge on only the new strand.
B)The incorrect base is in the parental strand which is modified with methyl groups.
C)Some bases of the parental strand are methylated, while none of the new strand are.
D)A methyl group is attached to the incorrect base in the newly synthesized DNA strand.
E)DNA polymerase is bound to the new strand at the location of the incorrect base.

Accepted Answer

The bacterial mismatch repair system distinguishes the newly synthesized strand from the parental strand based on methylation. In bacteria like E. coli, the parental DNA strand is methylated, while the newly synthesized strand is not immediately methylated. This difference allows the repair system to identify and correct mismatches on the new strand, ensuring that the original genetic information is preserved.

Question 4

What statement about biochemical pathways is correct?
A)All enzymes in the pathway catalyze the same reaction.
B)If an enzyme in a pathway is inactive, adding excessive amounts of its substrate will restore the normal phenotype.
C)If an enzyme in a pathway is inactive, adding excessive amounts of its product will restore the normal phenotype.
D)If the enzyme that catalyzes the final step in a pathway is inactive, all the other enzymes will be inactivated as well.
E)If the first enzyme in a pathway is inactivated, adding the final product will not restore the normal phenotype.

Accepted Answer

The answer of What statement about biochemical pathways is correct?
A)All...

Question 5

In the Ames test for mutagenicity,
A)reverse mutations convert auxotrophs to prototrophs that survive.
B)colonies are due to forward mutations that convert prototrophs to auxotrophs.
C)cells are treated with mutagen and only cells with no mutations survive.
D)cells are treated with excess amino acids and cells with mutations are killed.
E)rat liver enzymes are present to protect cells from mutation.

Accepted Answer

The Ames test detects reverse mutations where a substance causes genetic mutations that convert auxotrophic bacteria (which cannot grow without a specific nutrient supplementation) back to prototrophic bacteria (which can grow without the specific nutrient supplementation). This is used to assess the mutagenic potential of chemical compounds.

Question 6

A plaque is
A)a colony of bacteria growing on a plate.
B)a clump of bacterial cells that contain phage within them.
C)a region on a plate where bacteria that have survived phage infection are living.
D)an area on a plate that has live phage-resistant bacteria.
E)an area on a plate that has phage and dead bacteria.

Accepted Answer

A plaque is an area on a plate where phage has lysed bacterial cells, resulting in clear zones of dead bacteria.

Question 7

The first step in base excision repair is
A)removal of a double-stranded fragment of damaged DNA.
B)nicking the backbone of one DNA strand.
C)recognizing and excising the incorrect base from a nucleotide.
D)removing extraneous groups such as methyl or oxygen added by mutagens.
E)replacing an A-T base pair with a C-G base pair.

Accepted Answer

The first step in base excision repair involves recognizing and excising the incorrect or damaged base from a nucleotide. This is typically done by a specific glycosylase enzyme that identifies and removes the erroneous base, leaving behind an apurinic/apyrimidinic (AP) site.

Question 8

Fruit flies normally have red eyes.Seven different true-breeding strains of fly with white eyes have been identified (A-H).In each strain, the white eye trait is due to an autosomal recessive allele.It is possible that all seven strains have mutations in the same gene.Alternatively, some or all of the strains may have mutations in different genes.To determine how many genes are involved in eye color in these flies, all possible pairwise crosses are performed.The offspring phenotypes resulting from each cross are observed.(+ = wild-type; ? = mutant)
 $$\begin{array} { | c | c | c | c | c | c | c | c | c | } 
\hline & \mathrm { A } & \mathrm { B } & \mathrm { C } & \mathrm { D } & \mathrm { E } & \mathrm { F } & \mathrm { G } & \mathrm { H } \
\hline \mathrm { A } & - & - & + & + & - & + & + & - \
\hline \mathrm { B } & & - & + & + & - & + & + & - \
\hline \mathrm { C } & & & - & - & + & + & + & + \
\hline \mathrm { D } & & & & - & + & + & + & + \
\hline \mathrm { E } & & & & & - & + & + & - \
\hline \mathrm { F } & & & & & & - & - & + \
\hline \mathrm { G } & & & & & & & - & + \
\hline \mathrm { H } & & & & & & & & - \
\hline
\end{array}$$
-Based on these crosses, how many different eye color genes are mutant in all of these strains collectively?
A)1
B)2
C)3
D)4
E)8

Accepted Answer

The answer of Fruit flies normally have red eyes.Seven different...

Question 9

Which statement about amino acids is false?
A)A polypeptide is made up of several amino acids attached by peptide bonds.
B)Amino acids are linked by peptide bonds that form between two amino groups.
C)The C terminus of a polypeptide chain contains a free carboxylic acid group.
D)Each type of amino acid has a unique R group.

Accepted Answer

Amino acids are linked by peptide bonds that form between the carboxyl group of one amino acid and the amino group of another, not between two amino groups.

Question 10

How is the Ames test for mutagenicity useful for identifying potential carcinogens?
A)Bacteria do not get cancer so they can survive lethal carcinogens.
B)Substances that mutate bacterial DNA are likely to mutate human DNA.
C)Bacteria thrive on substances that could cause cancer in humans.
D)The same genes that cause cancer in humans can be mutated in bacteria.
E)Liver enzymes alter the bacteria so they will behave like human cells.

Accepted Answer

The Ames test is based on the principle that substances that can induce mutations in the DNA of bacteria (such as Salmonella typhimurium) may also have the potential to cause mutations in human DNA, which can lead to cancer. This test is a quick and cost-effective way to screen chemicals for mutagenic potential.

Question 11

What is true about the mutations within a complementation group?
A)They produce the same mutant phenotype.
B)They complement each other and are in the same gene.
C)They do not complement each other and are in the same gene.
D)They complement each other and are in different genes.
E)They do not complement each other and are in different genes.

Accepted Answer

Mutations within a complementation group do not complement each other because they affect the same gene, leading to the same mutant phenotype. This means that when two mutations are in the same gene, they cannot provide what the other lacks to restore the wild-type function, indicating they are part of the same complementation group.

Question 12

Experiment 2 depends on recombination.

Accepted Answer

Experiment 2, as described, likely involves genetic or molecular biology processes where recombination is a key mechanism for combining genetic material, which is essential for the experiment's objectives.

Question 13

In the human genetic disorder alkaptonuria, urine turns black because of the presence of homogentisic acid in individuals with the trait.This is due to
A)the presence of large amounts of homogentisic acid in the diet.
B)failure of individuals with alkaptonuria to manufacture enzymes involved in the synthesis of homogentisic acid.
C)failure of wild-type individuals to manufacture enzymes involved in the synthesis of homogentisic acid.
D)failure of the kidneys to remove homogentisic acid from the urine.
E)failure of individuals with alkaptonuria to manufacture enzymes involved in the breakdown of homogentisic acid.

Accepted Answer

Alkaptonuria is caused by the body's inability to produce enough of the enzyme needed to break down homogentisic acid, leading to its accumulation and excretion in urine, which turns black upon exposure to air.

Question 14

What is the consequence to a bacterial cell of a mutation that inactivates the enzyme responsible for methylating the A in the DNA sequence 5′ GATC?
A)Replication of the lagging strand cannot be completed.
B)Thymine dimers cannot be repaired.
C)Parental and new DNA strands cannot be distinguished during mismatch repair.
D)Expression of certain metabolic genes is decreased.
E)The mutation rate of all genes will be lower.

Accepted Answer

The enzyme responsible for methylating the A in the DNA sequence 5′ GATC is involved in distinguishing between the parental and newly synthesized DNA strands during mismatch repair. Methylation marks the parental strand, allowing the repair systems to correct mismatches in the newly synthesized strand by using the methylated strand as the template. If this enzyme is inactivated, the cell cannot differentiate between the old and new strands, impairing the mismatch repair process.

Question 15

Which statement about amino acids is false?
A)Every amino acid contains a carboxyl group.
B)The side chain or R group is the same in each amino acid.
C)In a polypeptide, amino acids are joined by peptide bonds.
D)One end of a polypeptide, the N terminus, contains a free amino group.

Accepted Answer

The side chain or R group varies among amino acids, giving each its unique properties.

Question 16

Consider the pathway for the synthesis of the amino acid arginine in Neurospora: $\begin{array}{ll}&A R G-E&&A R G-F&&A R G-H\
\text { ornithine }& \rightarrow &\text { citrulline } &\rightarrow& \text { argininosuccinate } &\rightarrow&\text { arginine }
\end{array}$
 Mutant strains of Neurospora may carry one or more mutations.Neurospora mutant strain b is grown on minimal media plus supplements as shown.Growth is shown by (+)and no growth is shown by (o). $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\underline{ Supplements }$
$\begin{array}{cccccc}
\begin{array}{c}
\text { mutant } \
\text { strain }
\end{array} & \text { nothing } & \text { ornithine } & \text { citrulline } & \begin{array}{c}
\text { arginino- } \
\text { succinate }
\end{array} & \text { arginine } \
\text { a } & 0 & 0 & 0 & + & + \
\text { b } & 0 & 0 & + & + & +
\end{array}$ What can you conclude from these data?
A)Strain a has only one mutation and it is in ARG-E.
B)Strain b has only one mutation and it is in ARG-H.
C)Strain a has a mutation in ARG-F and strain b has a mutation in ARG-E.
D)Strain b has mutations in ARG-E, ARG-F, and ARG-H.

Accepted Answer

Strain a grows with argininosuccinate and arginine, indicating a block before argininosuccinate, likely in ARG-F. Strain b grows with citrulline, argininosuccinate, and arginine, indicating a block before citrulline, likely in ARG-E.

Question 17

The genetic condition xeroderma pigmentosum, which can lead to skin cancer, results from the inability to
A)correct UV-induced thymine dimers.
B)process the amino acid phenylalanine.
C)produce functional hemoglobin.
D)correct transitions.
E)correct breaks in the X chromosome.

Accepted Answer

Xeroderma pigmentosum is a condition characterized by an extreme sensitivity to ultraviolet (UV) rays from sunlight. This sensitivity is due to the body's inability to repair damage to the DNA caused by UV light, specifically the inability to correct UV-induced thymine dimers. Thymine dimers form when two adjacent thymine bases bond together after being damaged by UV light, disrupting normal DNA replication and repair processes.

Question 18

Consider the pathway for the synthesis of the amino acid arginine in Neurospora: $\begin{array}{ll}&A R G-E&&A R G-F&&A R G-H\
\text { ornithine }& \rightarrow &\text { citrulline } &\rightarrow& \text { argininosuccinate } &\rightarrow&\text { arginine }
\end{array}$
 Mutant strains of Neurospora may carry one or more mutations.Neurospora mutant strain a is grown on minimal media plus supplements as shown.Growth is shown by (+)and no growth is shown by (o).Growing strain a cells accumulate citrulline. $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\text { Supplements }$
$\begin{array}{ccccc}
\begin{array}{c}
\text { mutant } \
\text { strain }
\end{array} & \text { nothing } & \text { ornithine } & \text { citrulline } &\begin{array}{c}
\text { arginino-}\
\text { succinate }
\end{array}&{ arginine }\
a&0&0&0&0&+
\end{array}$ Which statement is true about strain a?
A)It has one mutation in ARG-H.
B)It has two mutations, one in ARG-F and one in ARG-H.
C)It has two mutations, one in ARG-E and one in ARG-H.
D)It has two mutations, one in ARG-E and one in ARG-F.
E)It has mutations in ARG-E, ARG-F, and ARG-H.

Accepted Answer

The answer of Consider the pathway for the synthesis of...

Question 19

Which statement(s)about the rII- strain of T4 bacteriophage that Benzer studied is (are)true? (Select all that apply. )
A)It grows in E.coli B, but not in E.coli K(λ).
B)It produces larger plaques than wild-type.
C)It grows in E.coli K(λ), but not in E.coli B.
D)It produces smaller plaques than wild-type.

Accepted Answer

The answer of Which statement(s)about the rII- strain of T4...

Question 20

One strain of rII- phage has a deletion in the rII region, another has a point mutation in the rII region.When E.coli K(λ)cells are infected with either rII- phage (not both)no plaques form.When E.coli K(λ)cells are infected with both rII- phage simultaneously plaques do form.Why do plaques form when E.coli K(λ)cells are infected with both types of rII- phage simultaneously?
A)Recombination between the two phage genomes resulted in a wild-type phage.
B)The two phage have mutations in different rII genes and their genomes complement each other.
C)A protein made by the deletion phage can repair the DNA of the phage with the point mutation.
D)Bacteria can survive infection with one phage, but they cannot survive infection with two phages.

Accepted Answer

The two phage have mutations in different rII genes, allowing them to complement each other and produce functional proteins necessary for plaque formation.

Quiz 7: Anatomy and Function of a Gene: Dissection Through Mutation

Experiment 1 depends on recombination.

A bacterial mismatch repair system is able to correct replication errors that insert an incorrect nucleotide.How is this repair system able to determine which mismatched base is incorrect?

What statement about biochemical pathways is correct?

In the Ames test for mutagenicity,

A plaque is

The first step in base excision repair is

Which statement about amino acids is false?

How is the Ames test for mutagenicity useful for identifying potential carcinogens?

What is true about the mutations within a complementation group?

Experiment 2 depends on recombination.

In the human genetic disorder alkaptonuria, urine turns black because of the presence of homogentisic acid in individuals with the trait.This is due to

What is the consequence to a bacterial cell of a mutation that inactivates the enzyme responsible for methylating the A in the DNA sequence 5′ GATC?

Which statement about amino acids is false?

The genetic condition xeroderma pigmentosum, which can lead to skin cancer, results from the inability to

Which statement(s) about the rII- strain of T4 bacteriophage that Benzer studied is (are) true? (Select all that apply. )