Biology 1207 Midterm 4 Review

1. Which of the following choices will be affected by a cell containing two nonfunctional copies of p53? I. Apoptotic pathways II. DNA repair pathways III. Ability to arrest the cell cycle a. I and III b. I, II and III c. I and II d. II only

1. Which of the following choices will be affected by a cell containing two nonfunctional copies of p53? I. Apoptotic pathways II. DNA repair pathways III. Ability to arrest the cell cycle a. I and III b. I, II and III c. I and II d. II only

What is the inducer of the Lac Operon? Lac A Allolactose Lac I ATP

Allolactose

3. Predict the phenotype of a promoter (lac P) mutant which has a mutation in the promoter for the lac operon. a. The lac genes would be expressed efficiently only in the absence of lactose. b. The lac genes would be expressed efficiently only in the presence of lactose. c. The lac genes would be expressed continuously. d. The lac genes would never be expressed efficiently.

Be sure to be able to predict whether AND explain why the lac operon is expressed or not under the following conditions: no lactose, no glucose; no lactose, glucose present; both lactose and glucose present; lactose present but no glucose. This question requires that you understand the function and position of each of the components of the lac operon, how they interact with one another and control gene expression. Do you know how an inducer (Allolactose) affects the lac operon? Answer: Choice (d). Since the Lac P region is the promoter that binds RNA polymerase, a mutation that would block the function of this region would prevent the structural genes (Lac A, B and Y) of the lac operon from ever being expressed, regardless of the lactose conditions.

How is a new DNA strand synthesized in nucleotide excision repair? By crossing-over between sister chromatids By DNA polymerase By DNA ligase By Uvr A

By DNA polymerase

Which is one way a tumor suppressor gene could become mutated and lose function? Gene duplication Chromosome loss Insertion of active promoter infront of tumor suppressor gene

Chromosome loss

Heterochromatin allows transcription of genes tightly packed into nucleosomes. True False

F

Mice that do not have a functional p53 gene will be born with many abnormalities. True False

F

Mutations can only occur in the coding sequence of a gene. True False

F

The Mediator binds directly to the promoter in a eukaryotic gene. True False

F

Where is the TATA box located in a eukaryotic gene? In the upstream regulatory region In the core promoter In the coding region (part that contains exons and introns) At the termination site of the gene

In the core promoter

The structural genes in the Lac Operon are all dedicated to the breakdown of: Glucose Glycogen Lactose Sucrose

Lactose

A point mutation that results in an amino acid substitution is a __________ mutation. Nonsense Silent Missense

Missense

Which of the following is true concerning a somatic cell mutation? a. Half of the gametes carry the mutation. b. A small fraction of the gametes carry the mutation. c. Only a small group of cells within the organism is affected by the mutation. d. All cells within the organism are affected by the mutation. e. All of the gametes carry the mutation.

Only a small group of cells within the organism is affected by the mutation.

Repressors and activators affect the level of transcription in the cell. True False

T

The amplification of the Myc gene will lead to the over replication of cells because Myc is a transcription factor that activates target genes in the EGF signaling pathway. True False

T

The progression from one phase in the cell cycle to the next (e.g. G1 to S) is dependent on the increase in concentration of a cyclin-cdk complex. True False

T

Tumor suppressor genes are activated at checkpoints when DNA damage is detected. True False

T

If the cell's control of replication is like the brakes and accelerator of a car. An oncogene is similar to: The accelerator stuck in the "on" position The brakes of the car not working The accelerator not working The brakes of the car "on" all the time

The accelerator stuck in the "on" position

What part of the histone protein is modified to control nucleosome packing? a. The carboxyl end of the histone can be methylated, phosphorylated or acetylated b. The amino end of the histone can be methylated, phosphorylated or acetylated c. The entire length of the histone is covered with basic acetyl groups d. Segments of the histone are cleaved and removed so the DNA can be packed tighter around the nucleosome core.

The amino end of the histone can be methylated, phosphorylated or acetylated

What part of the histone protein is modified to control nucleosome packing? a. The carboxyl end of the histone can be methylated, phosphorylated or acetylated. b. The entire length of the histone is covered with basic acetyl groups. c. The amino end of the histone can be methylated, phosphorylated or acetylated. d. Segments of the histone are cleaved and removed so the DNA can be packed tighter around the nucleosome core.

The amino end of the histone can be methylated, phosphorylated or acetylated.

6. Histone Acetyltransferases are enzymes responsible for modifying histone structure. These enzymes add acetyl groups to amino acids on the histone protein that contain amine groups on their side chains (arginine and lysine). By adding acetyl groups to these amino acid side chains, the structure of the entire histone protein complex will change, thereby altering the expression of different genes associated with these histones. If histone X were to be acetylated, the expression of gene Y would be increased. If gene Y codes for protein Z, what would happen to the levels of protein Z if the concentration of histone acetylase were to decrease? a. The concentration of protein Z would increase, as the expression of gene Y is increasing. b. The concentration of protein Z would decrease, as the expression of gene Y is decreasing. c. The concentration of protein Z would increase, as the expression of gene Y is decreasing. d. The concentration of protein Z would decrease, as the expression of gene Y is increasing.

This question requires an understanding of the effect histone modifying enzymes have on gene expression. Most methylation of histone amino acids results in repressing transcription, while acetylation of certain histone amino acids result in the activation of transcription. Answer: Choice (b). The production of protein Z is dependent upon gene Y. Acetylation of the histones associated with gene Y will promote transcription of gene Y and thereby increase the production of protein Z. If acetylation of histones associated with gene Y would decrease, then the production of protein Z would also decrease.

2. A non-conservative mutation is referred to as one that results in the exchange of a single, new amino acid that has different biochemical properties for the original one in the polypeptide. Which one of these could be a non-conservative mutation? a. Silent b. Missense c. Nonsense d. Frameshift

This question requires that you understand each of these four types of mutations, how they are different from one another, and what effect they have on the polypeptide product. Answer: Choice (b). The definition of a missense mutation is when one amino acid is switched for a different amino acid. A silent mutation results in no amino acid change, a nonsense mutation substitutes a stop codon for the codon calling for an amino acid, and a frameshift mutation inserts a base in the code, so that the reading frame for all the codons that follow has changed. This results in a complete change in many amino acids.

4. Which of the following will least likely affect the length of a protein product? a. Nonsense mutation b. Single-base deletion c. Missense mutation d. Frameshift mutation

This question requires that you understand the effect each of these types of mutations have on the resulting polypeptide. Answer: Choice (c) because a single amino acid is substituted for another in a missense mutation and this doesn't change the length of the protein. In choice a, the nonsense mutation will lead to a shortened protein (introduces an early stop codon, in choice b, a single-base deletion results in a frame-shift and that will often lead to an early stop codon, and choice d, as just mentioned, often leads to an early stop codon.

1. All the following mutations can result in a reduction of β-galactosidase synthesis (in the Lac Operon) except: a. A mutation in adenylate cyclase b. A mutation in catabolite activator protein (CAP). c. A mutation in the CAP site in the lac control region. d. A mutation in the repressor binding site in the operator.

This question requires that you understand the function and position of each of the components of the lac operon, how they interact with one another and control gene expression. Answer: Choice (d). A mutation in the Lac O region, binding site of the repressor, would not reduce the amount of B-gal because binding of the Lac repressor to Lac O normally blocks the synthesis of B-gal. All the other choices would prevent or reduce the synthesis of B-gal because the production and binding of the CAP protein (bound to cAMP made by adenyl cyclase) would be compromised by these mutations.

4. Predict the phenotype of an operator mutant (Oc) which prevents the binding of the repressor to lac O (the operator), where "c" means that the operator cannot bind to a repressor protein. a. The lac genes would be expressed efficiently only in the absence of lactose. b. The lac genes would be expressed efficiently only in the presence of lactose. c. The lac genes would be expressed continuously, sometimes at a basal level. d. The lac genes would never be expressed efficiently.

This question requires that you understand the function and position of each of the components of the lac operon, how they interact with one another and control gene expression. Mutations in the lac operon were critical in defining the role of each gene in the lac operon. It is important that you know which genes are regulatory (lac P = promoter, lac O = operator, binding site of lac repressor) and those that actually encode for proteins (lac Z = B-galactosidase that breaks lactose into glucose and galactose, lac Y = lactose permease that is a transporter that brings lactose into the bacterial cell, and lac A = B-galactoside transacetylase that modifies lactose. Answer: Choice (c). The whole purpose of the Lac repressor is to allow the lac operon to be repressed when it is not needed, when there is no lactose to be metabolized into glucose. Since a mutation in the operator (binding site of the Lac repressor) would prevent the shutting down of the lac operon in the absence of lactose, the lac genes would be expressed continuously. The lac operator is situated between the lac promoter and the structural genes of the lac operon to block the expression of the lac operon genes in the absence of lactose. If this repression doesn't take place, the lac structural genes are free to be expressed, even when there is no lactose to be metabolized into glucose.

5. A point mutation within which functional part of a DNA sequence would be most likely to ultimately result in the production of proteins which differ from the non-mutated form by only a single amino acid? a. Exon b. Promoter c. Intron d. Telomere e. Any choice could be correct.

This question requires that you understand the function of each of these parts of the gene and which encodes amino acid sequence. What if the mutation didn't change the amino acid sequence but just affected the level of the protein in the cell? Answer: Choice (a), which is the only choice given that contains DNA sequence that codes for amino acid sequence. All the other choices are either not found in the mature/processed mRNA (choice b and choice c) or are a part of the chromosome that is never transcribed (choice d).

2. Which of the following is true about the DNA-binding and activation domains of transcription activators: a. The DNA-binding domain is always at the N-terminus, and the activation domain is at the C-terminus. b. The activation domain of one protein can be fused to the DNA-binding domain of another protein to generate a functional activator protein. c. The DNA-binding domain is always at the C-terminus, and the activation domain is at the N-terminus. d. The distance between a DNA-binding domain and activation domain cannot be altered.

This question requires that you understand the role of DNA-binding domains and activation or repression domains in regulatory protein. An activator is comprised of two separate domains within the same protein: a DNA-binding domain that directs this protein to bind to a certain DNA sequence and an activation domain that activates the transcription of the gene that the protein has bound to. One controls guiding the protein to its site in the genome, and the other domain executes the actual function, to activate the expression of the gene. The same is true for a repressor protein, only it has a domain that executes the repression of the target gene's expression when it binds. Both of these proteins have to have DNA-binding domains, or they cannot affect the transcription of their target genes. These domains (DNA-binding and activation/repression are interchangeable). For example, if a protein normally represses gene A, but has had its repressor domain replaced with an activator domain, now the same protein will activate gene A. Answer: Choice (b). The distance between a DNA-binding domain and an activation domain can be variable and these domains are not required to be near either end of the protein. Hybrid proteins are made routinely by fusing the coding region of the DNA-binding domain of one gene with the activation domain of another gene. This creates a protein that will bind to the DNA corresponding to the DNA-binding domain, while activating the expression of the target gene.

3. Which of the following represents a frameshift mutation of the given template strand (with the codons indicated) here: 5'-AGC-CTT-AGC-3' a. 5'-AGC-CTT-AGG-3' b. 5'-AGC-GCT-TAG-C-3' c. 5'-TTT-AGC-CTT-AGC-3' d. 5'-TGC-CTT-AGC-3' e. 5'-CTT-AGC-3'

This question requires that you understand what a frameshift mutation is and how to recognize it in a changed nucleotide sequence. Answer: Choice (b). This question requires that you understand that a frameshift mutation is the introduction of a base (or deletion of a base) that changes the reading frame or codons of the remaining sequence. In choice a, the last base in the sequence is changed from a C to a G, so this is not a frameshift mutation. In choice c, three T's are added to the 5' end of the template, so this would not change the reading frame because it is an insertion of three bases. In choice d, the first base is changed from A to T, so like choice a, this is not a frameshift mutation. Finally, choice e deletes the first three bases of the sequence and this itself doesn't change the reading frame of the sequence.

6. During which phase of the cell cycle are DNA repair mechanisms least active? a. G1 b. S c. G2 d. M

This question requires that you understand what is taking place in each of the phases of the cell cycle and when DNA repair mechanisms take place. [There is more than one way to repair DNA.] Answer: Choice d, because during Mitosis, the chromosomes are condensed, align during metaphase and move to opposite daughter cells during anaphase. There is no time for DNA repair during M phase, but DNA repair must take place during the checkpoints in G1, G2 and by DNA polymerase during the S phase (when the DNA is being synthesized).

5. You have discovered a mutation in a gene that completely shuts down the initiation of transcription. You look at the sequence of the gene and find that the mutation in the: a. TATA box b. 5' splice site between the first exon and intron c. Termination of transcription sequence d. 5' cap site

This question tests your understanding of the different parts of the gene and the role certain parts of the gene play in regulating transcription. Answer: Choice (a). Initiation of transcription is focused in the promoter region of a gene. The only choices that are relevant to the initiation of transcription is the TATA box (choice a). The other choices relate to steps that occur after the initiation of transcription such as processing the mRNA (choice b), ending transcription (choice c) and processing the mRNA (choice d).

Thymine Dimers are caused by: Chemical mutagens UV irradiation Smoking Spontaneous errors in DNA replication

UV irradiation

The likely outcome from a mutation in the lacO site (lac Operator) of the lac operon would be a. the order in which the genes of the lac operon are transcribed could be altered. b. binding of the activator protein could be hindered. duplication could be affected. c. binding of RNA polymerase could be hindered. d. binding of a repressor protein could be hindered.

binding of a repressor protein could be hindered.

How is a new DNA strand made in nucleotide excision repair? by crossing over between sister chromatids by DNA polymerase by DNA ligase by DNA exonuclease

by DNA polymerase

Liver cells, mammary cells, and skin cells all contain the same genome; however, their respective proteomes vary drastically. This observation is best explained by what phenomenon? a. evolution b. cell division c. cell differentiation d. crossing over

cell differentiation

The best description of the role of the mediator protein complex in eukaryotes is it a. is composed of a single protein. b. facilitates interactions between RNA polymerase II and regulatory transcription factors. c. controls the rate at which RNA polymerase translates mRNA. d. completely enfolds RNA polymerase and the general transcription factors.

facilitates interactions between RNA polymerase II and regulatory transcription factors.

The chemical, EMS (ethyl methanesulfonate) can cause ____________ mutations. induced spontaneous

induced

Which statement is TRUE regarding the effect of alternative splicing of pre-mRNAs? a. it increases the size of both the genome and the proteome b. it has no effect on the genome, while increasing the size of the proteome c. it has no effect on the proteome, while increasing the size of the genome d. it decreases the size of both the genome and the proteome e. it increases the size of both the genome while decreasing the proteome

it has no effect on the genome, while increasing the size of the proteome

When glucose is high, cAMP is _____________. high low doesn't change

low

How does a mutation in p21 Ras result in an oncogene? p21 Ras normally encodes a G-protein that represses cell division. p21 Ras normally encodes a G-protein that promotes cell division.

p21 Ras normally encodes a G-protein that represses cell division.