Chapter 14, 15, 16, and 17 LEAP questions

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

A

Hemophilia is an X-linked recessive condition in which blood does not clot properly. Queen Victoria of England had one allele for hemophilia. Most of her male descendants had the disorder, but few females had it. Why did hemophilia occur more frequently in Queen Victoria's male descendants? a. Only one copy of the X chromosome is found in cells of males, but two copies are found in cells of females. b. Replication of the X chromosome occurs often in males but rarely in females. c. Males have hormones that enhance the expression of X-linked traits. d. Males mature more slowly than females, allowing recessive traits more time to appear.

A

In Drosophila flies, eye color is sex-linked. The allele for red eyes (XW) is dominant to the allele for white eyes (Xw). A male fly with red eyes XW Y and a carrier (heterozygote) female XW Xw are mated. What is the probability that the female offspring will have white eyes? a. 0% b. 100% c. 25% d. 75% e. 50%

A

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 is in the: a. TATA box b. 5' splice site between the first exon and intron c. Termination of transcription sequence d. 5' cap site

A

A cell has 12 pairs of homologous chromosomes. How many chromosomes does each daughter cell have after the completion of Meiosis I? a. 6 b. 12 c. 24 d. 48

B

A cell has 12 pairs of homologous chromosomes. How many chromosomes does each daughter cell have after the completion of Meiosis II? a. 6 b. 12 c. 24 d. 48

B

A cell isolated from a tumor was studied and found to rapidly advance from G1 to S phase. This continued despite exposing the cell to UV irradiation. The irradiated cell continued to divide rapidly and passed through the G1 to S phase transition quickly. What aspect of cell cycle regulation could have been abnormal in this cell? a. Lack of production of cyclins b. Faulty G1 checkpoint c. Overexpression of p53 d. Lack of production of cyclin-dependent kinases e. All of these choices could be responsible for the rapidly dividing tumor cell after UV irradiation

B

A cell that is dividing lacks the ability to make kinetochore microtubules while it is still able to synthesize and use astral and polar microtubules and microfilaments. During which stage of Mitosis will this defect first be visibly evident under the microscope? a. Prophase b. Metaphase c. Anaphase d. Telophase e. Cytokinesis

B

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

B

Aneuploidy is the presence of an abnormal number of chromosomes in a cell. Monosomy is a subtractive form of aneuploidy due to the presence of only one chromosome from a pair while trisomy is an additive from where there are three instances of a particular chromosome, instead of the normal two. A strawberry plant is normally diploid, but a monosomic strawberry plant was observed with 9 chromosomes in its cells. Which of the following is NOT a correct possibility regarding the chromosomal material of a strawberry plant? a. A trisomic strawberry plant would have 11 chromosomes in its cells b. A trisomic strawberry plant would have 15 chromosomes in its cells c. Doubling the chromosomal number found in the normal haploid state would result in a strawberry plant with 10 chromosomes in its cells During meiosis of the normal diploid, the chromosomal number is halved during the first meiotic division, yielding 5 chromosomes

B

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

B

Which is an example of aneuploidy in a human? a. A male somatic cell has one X chromosome and one Y chromosome. b. A female gamete contains two X chromosomes. c. A female somatic cell contains 46 chromosomes. d. A male gamete has only one Y-chromosome.

B

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

B

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

B

Which of the following represents a frameshift mutation of the given template strand (with the codons indicated): 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'

B

A cell has 12 pairs of homologous chromosomes. How many chromosomes does it have in Metaphase of Mitosis? a. 6 b. 12 c. 24 d. 48

C

Consider a diploid species where n=6, where n is the number of chromosomes in a haploid gamete. If an individual of this species was found to have 18 chromosomes, it would be categorized as: a. Aneuploid b. Tetraploid c. Polyploid d. Monosomic e. Trisomy

C

In humans, hair texture exhibits incomplete dominance. The gene for curly hair (H) is incompletely dominant to the gene for straight hair (h). Individuals that are heterozygous (Hh) have wavy hair. Two heterozygous parents have a child. What is the chance that the child will have wavy hair? a. 0% b. 25% c. 50% d. 75% e. 100%

C

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

C

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

C

All of the following mutations can result in a reduction of beta-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

D

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

D

If a cell with 4 pairs of chromosomes, skips Meiosis II while making a total of two gametes, how many total chromosomes and chromatids will each gamete have at the end of this abnormal Meiosis? a. 8 chromosomes and 8 chromatids b. 8 chromosomes and 4 chromatids c. 4 chromosomes and 4 chromatids d. 4 chromosomes and 8 chromatids

D

In some chickens, feather color is controlled by codominance. When a clack feathered chicken mates with a white feathered chicken, all of the offspring are covered in both black and white feathers. A farmer mates a black feathered chicken (BB) with a black-and-white feathered (BW) chicken. What are the predicted phenotypes of their offspring? a. All of the offspring will have black feathers. b. All of the offspring will have black-and-white feathers. c. 75% of the offspring will be black feathered, and 25% of the offspring will be white feathered. d. 50% of the offspring will be black feathered and 50% of the offspring will be black-and-white feathered.

D

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

D