Bio Unit 6 Questions Part 2

Antigens are foreign proteins that invade the systems of organisms. Vaccines function by stimulating an organism's immune system to develop antibodies against a particular antigen. Developing a vaccine involves producing an antigen that can be introduced into the organism being vaccinated and which will trigger an immune response without causing the disease associated with the antigen. Certain strains of bacteria can be used to produce antigens used in vaccines. Which of the following best explains how bacteria can be genetically engineered to produce a desired antigen? A The gene coding for the antigen can be inserted into plasmids that can be used to transform the bacteria. B The bacteria need to be exposed to the antigen so they can produce the antibodies. C The DNADNA of the antigen has to be transcribed in order for the mRNAmRNA produced to be inserted into the bacteria. D The mRNAmRNA of the antigen has to be translated in order for the protein to be inserted into the bacteria.

A

Arsenic is a toxic element found in both aquatic and terrestrial environments. Scientists have found genes that allow bacteria to remove arsenic from their cytoplasm. Arsenic enters cells as arsenate that must be converted to arsenite to leave cells. Figure 1 provides a summary of the arsenic resistance genes found in the operons of three different bacteria. E. coli R773R773 is found in environments with low arsenic levels. Herminiimonas arsenicoxydans and Ochrobactrum tritici are both found in arsenic‑rich environments Researchers claim that bacteria that live in environments heavily contaminated with arsenic are more efficient at processing arsenic into arsenite and removing this toxin from their cells. Justify this claim based on the evidence shown in Figure 1. A There are multiple operons controlling the production of proteins that process and remove arsenite from cells in both H. arsenicoxydans and O. tritici. In contrast, E. coli has only one operon devoted to arsenic removal. B Both H. arsenicoxydans and O. tritici contain the arsR���� gene that codes for a repressor that turns on the operon to eliminate arsenite from the cell. C Both O. tritici and E. coli contain the arsD���� gene, which codes for a protein that helps remove arsenite from the cell. D Both H. arsenicoxydans and O. tritici. have more arsenic resistance genes than has E. coli.

A

Both liver cells and lens cells have the genes for making the proteins albumin and crystalline. However, only liver cells express the blood protein albumin and only lens cells express crystalline, the main protein in the lens of the eye. Both of these genes have enhancer sequences associated with them. The claim that gene regulation results in differential gene expression and influences cellular products (albumin or crystalline) is best supported by evidence in which of the following statements? A Liver cells possess transcriptional activators that are different from those of lens cells. B Liver cells and lens cells use different RNARNA polymerase enzymes to transcribe DNADNA. C Liver cells and lens cells possess the same transcriptional activators. D Liver cells and lens cells possess different general transcription factors.

A

Phytochromes are molecules that change light stimuli into chemical signals, and they are thought to target light-activated genes in plants. A study was conducted to determine how certain cell proteins were made in a plant cell using a phytochrome. Figures 1 and 2 represent findings from the study. Use the response models shown in Figures 1 and 2 to justify the claim that phytochromes regulate the transcription of genes leading to the production of certain cellular proteins. A When inactive phytochrome PrPr is activated by red light to become phytochrome PfrPfr, it is transported into the nucleus where it binds to the transcription factor PIF3PIF3 at the promoter. This stimulates transcription, ultimately leading to protein production. Far-red light inactivates the phytochrome, which will turn transcription off by not binding to PIF3PIF3. B Far-red light activates phytochrome PrPr, causing it to travel to the nucleus where it binds to PIF3PIF3 at the promoter. This stimulates transcription, ultimately leading to protein production. Red light inactivates the phytochrome, which will turn transcription off by not binding to PIF3PIF3. C MYBMYB, and not PfrPfr, is activated by red light, causing it to bind to the promoter and stimulate transcription and translation of cellular proteins. D PIF3PIF3 binds to the promoter only in the presence of red light and PfrPfr. Any time PIF3PIF3 is bound to the promoter, MYBMYB is transcribed, initiating transcription of various other proteins in the cell.

A

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 CAPCAP site prevents the binding of RNARNA polymerase when glucose is present in the environment. The lacZ, lacY, and lacA genes code for proteins needed for lactose metabolism. 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.

B

Nondisjunction during meiosis can negatively affect gamete formation. A model showing a possible nondisjunction event and its impact on gamete formation is shown in Figure 1. Which of the following best describes the most likely impact on an individual produced from fertilization between one of the daughter cells shown and a normal gamete? A Because nondisjunction occurred in anaphase II, all gametes will be normal and the resulting individual will be phenotypically normal. B Because nondisjunction occurred in anaphase II, all gametes will have an abnormal chromosome number and the individual will likely exhibit phenotypic evidence of the nondisjunction event. C Because nondisjunction occurred in anaphase IIII, all gametes will be normal and the resulting individual will be phenotypically normal. D Because nondisjunction occurred in anaphase IIII, all gametes will have an abnormal chromosome number and the individual will likely exhibit phenotypic evidence of the nondisjunction event.

B

The enzyme lactase aids in the digestion of lactose, a sugar found in the milk of most mammals. In most mammal species, adults do not produce lactase. Continuing to produce lactase into adulthood in people is called lactase persistence. A number of different alleles have been identified that result in lactase persistence. Figure 1 shows the percentage of people in different geographic areas parts of the Old World that exhibit lactase persistence. Which of the following best explains the distribution of lactase persistence in the areas shown in Figure 1 ? A Lactase persistence developed because people were malnourished in Europe. B Lactase persistence alleles are present in all human populations and are expressed when lactose is consumed. C Mutations conferring lactase persistence likely arose independently in different geographic areas and offered a selective advantage. D The mutations that cause lactase persistence are detrimental to humans and will eventually disappear from the gene pool.

C

Which of the following statements best explains the pattern seen on the gel with regard to the size and charge of molecules AA and BB? A Molecules AA and BB are positively charged, and molecule AA is smaller than molecule BB. B Molecules AA and BB are positively charged, and molecule AA is larger than molecule BB. C Molecules AA and BB are negatively charged, and molecule AA is smaller than molecule BB. D Molecules AA and BB are negatively charged, and molecule AA is larger than molecule BB.

C

ickle-cell anemia is an inherited blood disorder in which one of the hemoglobin subunits is replaced with a different form of hemoglobin. Partial DNADNA sequences of the HBB��� gene for normal hemoglobin and for sickle-cell hemoglobin are shown in Figure 1. Which of the following best describes the type of mutation shown in Figure 1 that leads to sickle-cell anemia? A Insertion B Deletion C Substitution D Frameshift

C

Genetic engineering techniques can be used when analyzing and manipulating DNADNA and RNARNA. Scientists used gel electrophoresis to study transcription of gene L� and discovered that mRNAmRNA strands of three different lengths are consistently produced. Which of the following explanations best accounts for this experimental result? A Gel electrophoresis can only be used with DNADNA (not mRNAmRNA), so experimental results are not interpretable. B RNARNA polymerase consistently makes the same errors during transcription of gene L�. C Gene L� is mutated, so RNARNA polymerase does not always transcribe the correct sequence. D Pre-mRNAmRNA of gene L� is subject to alternative splicing, so three mRNAmRNA sequences are possible.

D

Histone methyltransferases are a class of enzymes that methylate certain amino acid sequences in histone proteins. A research team found that transcription of gene R� decreases when histone methyltransferase activity is inhibited. Which scientific claim is most consistent with these findings? A DNADNA methylation inhibits transcription of gene R�. B Histone modifications of genes are usually not reversible. C Histone methylation condenses the chromatin at gene R� so transcription factors cannot bind to DNADNA. D Histone methylation opens up chromatin at gene R� so transcription factors can bind to DNADNA more easily.

D

Which of the following scientific claims is most consistent with the information provided in Figure 1 ? A Gene X� codes for a transcription factor required for transcription of gene D�. B A single transcription factor regulates transcription similarly, regardless of the specific gene. C Transcription of genes A�, B�, and C� is necessary to transcribe gene E�. D Different genes may be regulated by the same transcription factor.

D