A.P. Biology 2nd Trimester Exam (MC)

Enzyme used to position nucleotides during DNA replication (A)DNA ligase (B)DNA polymerase (C)RNA polymerase (D)Restriction enzyme (E)Reverse transcriptase

DNA polymerase

In a transformation experiment, a sample of E. coli bacteria was mixed with a plasmid containing the gene for resistance to the antibiotic ampicillin (ampr). Plasmid was not added to a second sample. Samples were plated on nutrient agar plates, some of which were supplemented with the antibiotic ampicillin. The results of E. coli growth are summarized below. The shaded area represents extensive growth of bacteria; dots represent individual colonies of bacteria. I) No AMP- lawn III) No AMP/Plasmid- lawn II) AMP- none IV) AMP/Plasmid- speckled Plates that have only ampicillin-resistant bacteria growing include which of the following? (A)I only (B)III only (C)IV only (D)I and II

IV only

The Trp operon is a coordinately regulated group of genes (trpA-trpE) that are required for tryptophan biosynthesis in E. coli. Based on the figure above, which of the following correctly describes the regulation of the Trp operon? (A)In the absence of tryptophan, the repressor is active and binds to the Trp operator, preventing RNA polymerase from transcribing the operon. (B)In the presence of tryptophan, the repressor is active and binds to the Trp operator, preventing RNA polymerase from transcribing the operon. (C)In the absence of tryptophan, the trpR gene is inactive, preventing the production of the repressor that blocks expression of the operon. (D)In the presence of tryptophan, the trpR gene is inactive, preventing the production of the repressor that blocks expression of the operon.

In the presence of tryptophan, the repressor is active and binds to the Trp operator, preventing RNA polymerase from transcribing the operon.

In a transformation experiment, a sample of E. coli bacteria was mixed with a plasmid containing the gene for resistance to the antibiotic ampicillin (ampr). Plasmid was not added to a second sample. Samples were plated on nutrient agar plates, some of which were supplemented with the antibiotic ampicillin. The results of E. coli growth are summarized below. The shaded area represents extensive growth of bacteria; dots represent individual colonies of bacteria. I) No AMP- lawn III) No AMP/Plasmid- lawn II) AMP- none IV) AMP/Plasmid- speckled Which of the following statements best explains why there are fewer colonies on plate IV than on plate III? (A)Plate IV is the positive control. (B)Not all E. coli cells are successfully transformed. (C)The bacteria on plate III did not mutate. (D)The plasmid inhibits E. coli growth.

Not all E. coli cells are successfully transformed.

Which of the following statements best explains the structure and importance of plasmids to prokaryotes? (A)Plasmids are circular, single-stranded RNARNA molecules that transfer information from the prokaryotic chromosome to the ribosomes during protein synthesis. (B)Plasmids are circular, double-stranded DNADNA molecules that provide genes that may aid in survival of the prokaryotic cell. (C)Plasmids are single-stranded DNADNA molecules, which are replicated from the prokaryotic chromosome, that prevent viral reproduction within the prokaryotic cell. (D)Plasmids are double-stranded RNARNA molecules that are transmitted by conjugation that enable other prokaryotic cells to acquire useful genes.

Plasmids are circular, double-stranded DNADNA molecules that provide genes that may aid in survival of the prokaryotic cell.

Enzyme used in the synthesis of mRNA (A)DNA ligase (B)DNA polymerase (C)RNA polymerase (D)Restriction enzyme (E)Reverse transcriptase

RNA polymerase

Enzyme found in retroviruses that produce DNA from an RNA template (A)DNA ligase (B)DNA polymerase (C)RNA polymerase (D)Restriction enzyme (E)Reverse transcriptase

Reverse transcriptase

For the following group of questions first, study the description of the situation or data and then choose the one best answer to each question following it and fill in the corresponding oval on the answer sheet. The diagram below illustrates the results of electrophoresis of DNA sequences obtained from a family of two adults and three children, and amplified using PCR. The bands represent short repeating sequences of variable length. Results for another female (X) are included for comparison. Father- 1, 2, 3 Mother- 1, 2, X Child 1- father, mother, 2, X Child 2- mother, father, 1, 3 Child 3- 2, father X- mother, 1, 2 Which of the following is the best explanation for the fragment pattern for individual X ? (A)She has only one member of this chromosome pair. (B)She has only one living parent. (C)She is homozygous for this particular DNA fragment. (D)She is the mother's child from another marriage. (E)She is not related to any member of the family being tested.

She is the mother's child from another marriage.

Erwin Chargaff investigated the nucleotide composition of DNA. He analyzed DNA from various organisms and measured the relative amounts of adenine (A), guanine (G), cytosine (C), and thymine (T) present in the DNA of each organism. Table 1 contains a selected data set of his results. Table 1. Nucleotide composition of sample DNA from selected organisms Which of the following statements best explains the data set? (A)Since the %A%A and the %G%G add up to approximately 50 percent in each sample, adenine and guanine molecules must pair up in a double-stranded DNADNA molecule. (B)Since the %A%A and the %T%T are approximately the same in each sample, adenine and thymine molecules must pair up in a double-stranded DNADNA molecule. (C)Since the %(A+T)%(A+T) is greater than the %(G+C)%(G+C) in each sample, DNADNA molecules must have a poly-AA tail at one end. (D)Since the %C%C and the %T%T add up to approximately 50 percent in each sample, cytosine and thymine molecules must both contain a single ring.

Since the %A and the %T are approximately the same in each sample, adenine and thymine molecules must pair up in a double-stranded DNADNA molecule. The percentages of adenine and thymine are approximately the same because adenine aligns with thymine in double-stranded DNADNA and they are therefore present in each organism in the same amounts.

Antibiotics can be used to kill the specific pathogenic bacterium, Mycobacterium tuberculosis, that causes tuberculosis. The appearance of antibiotic-resistant strains has made it more difficult to cure M. tuberculosis infections. These antibiotic-resistant bacteria survive and pass on the genes to their offspring, making the resistant phenotype more common in the population. DNA analysis indicates that the genes for antibiotic resistance are not normally present in bacterial chromosomal DNA. Which of the following statements best explains how the genes for antibiotic resistance can be transmitted between bacteria without the exchange of bacterial chromosomal DNA? (A)The antibiotic-resistant bacteria release a hormone that signals neighboring bacteria to become resistant. (B)The genes for antibiotic resistance are located on a plasmid that can be passed to neighboring bacteria. (C)The antibiotic-resistant bacteria are the result of bacteria that specifically modify their own chromosomal DNADNA to neutralize the antibiotics. (D)The antibiotic alters the bacterial genome of each bacterium, which results in an antibiotic-resistant population.

The genes for antibiotic resistance are located on a plasmid that can be passed to neighboring bacteria. Bacteria can exchange and express the genes found on plasmids, which are foreign, extrachromosomal loops of DNADNA that they pick up. Genes for antibiotic resistance are located on these plasmids.

In a transformation experiment, a sample of E. coli bacteria was mixed with a plasmid containing the gene for resistance to the antibiotic ampicillin (ampr). Plasmid was not added to a second sample. Samples were plated on nutrient agar plates, some of which were supplemented with the antibiotic ampicillin. The results of E. coli growth are summarized below. The shaded area represents extensive growth of bacteria; dots represent individual colonies of bacteria. I) No AMP- lawn III) No AMP/Plasmid- lawn II) AMP- none IV) AMP/Plasmid- speckled Which of the following best explains why there is no growth on plate II? (A)The initial E. coli culture was not ampicillin-resistant. (B)The transformation procedure killed the bacteria. (C)Nutrient agar inhibits E. coli growth. (D)The bacteria on the plate were transformed.

The initial E. coli culture was not ampicillin-resistant.

The following the DNA sequence is a small part of the coding (nontemplate) strand from the open reading frame of β-hemoglobin gene. Given the codon chart listed below, what would be the effect of a mutation that deletes the G at the beginning of the DNA sequence? 5'- GTT TGT CTG TGG TAC CAC GTG GAC TGA - 3' (A)The mutation precedes the gene, so no changes would occur. (B)Lysine (lys) would replace glutamine (gln), but there would be no other changes. (C)The first amino acid would be missing, but there would be no other change to the protein. (D)The reading frame of the sequence would shift, causing a change in the amino acid sequence after that point.

The reading frame of the sequence would shift, causing a change in the amino acid sequence after that point.

Figure 1 represents a metabolic process involving the regulation of lactose metabolism by E. coli bacteria. Lactose is utilized for energy by E. coli when glucose is not present. Allolactose is an isomer of lactose that is in the environment of these bacteria when lactose is present. The CAP site prevents the binding of RNA polymerase when glucose is present in the environment. The lacZ, lacY, and lacA genes code for proteins needed for lactose metabolism. Figure 1. Model of lac operon, comparing repressed and active states Which is a scientific claim that is consistent with the information provided and Figure 1 ? (A)The presence of excess lactose blocks the functioning of RNARNA polymerase in this operon. (B)When bound to the operator, the repressor protein prevents lactose metabolism in E. coli. (C)The binding of the repressor protein to the operator enables E. coli to metabolize lactose. (D)Allolactose acts as an inducer that binds to the operator, allowing E. coli to metabolize lactose.

When bound to the operator, the repressor protein prevents lactose metabolism in E. coli. When the repressor protein is bound to the operator, the lac operon is turned off and the genes for lactose metabolism cannot be transcribed.

For the following group of questions first, study the description of the situation or data and then choose the one best answer to each question following it and fill in the corresponding oval on the answer sheet. The diagram below illustrates the results of electrophoresis of DNA sequences obtained from a family of two adults and three children, and amplified using PCR. The bands represent short repeating sequences of variable length. Results for another female (X) are included for comparison. Father- 1, 2, 3 Mother- 1, 2, X Child 1- father, mother, 2, X Child 2- mother, father, 1, 3 Child 3- 2, father X- mother, 1, 2 The banding patterns of the DNA fragments reveal that (A)child 1 and child 2 cannot be biological siblings (B)child 1 and child 3 probably look like the mother (C)the mother cannot be the biological parent of all three children (D)the mother's DNA has the same DNA sequence as the father's DNA (E)child 2 and child 3 inherited all of their DNA from the father

the mother cannot be the biological parent of all three children