Follicular lymphoma and bcl-2
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Follicular lymphoma and bcl-2
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[post_date] => 2024-12-23 18:01:06
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[post_content] => Practice Passage (Question 1-6)
*This passage is the property of Khan Academy and has been reformatted into an AAMC-style interface in their entirety by MedLife Mastery. MedLife Mastery does not endorse and is not an affiliate of Khan Academy.
Follicular lymphoma is a slow progressing cancer that involves the clustering of B cells in the lymph nodes. B cells in an individual with follicular lymphoma do not divide much faster than normal B cells; however, they have an unusually long lifespan – much longer than that of normal B cells. Scientists believe this extended lifespan involves changes in chromosomes 14 and 18, which appear differently in the longer-living B cells than they do in normal B cells. In many cases of follicular lymphoma, there is a translocation between chromosomes 14 and 18, which leads to excess expression of the bcl-2 protein. Scientists hypothesize that the over-expression of bcl-2 might be related to the extended lifespan of these B cells.
Researchers testing this hypothesis created sense and antisense phosphorothioate (PS) oligodeoxynucleotides of the bcl-2 gene. A PS oligodeoxynucleotide is similar to a phosphodiester oligodeoxynucleotide (i.e. DNA), except that an oxygen atom in the backbone has been replaced by a sulfur atom (Figure 1), which increases the stability of the molecule. The researchers observed cell density over time when the PS oligodeoxynucleotides were introduced to colonies of B cells that expressed bcl-2 in excess. The results are shown in Figure 2.
Figure 1 Structure of a PS oligodeoxynucleotide
Figure 2 Cell density versus time in experimental and control groups
[post_title] => Follicular lymphoma and bcl-2
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[question] => What is the most likely reason for the change in bcl-2 expression in B cells in cases of follicular lymphoma?
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[answer] => 3
[description] => Reason for Correct Answer:
Translocations can influence gene expression in a variety of ways, including:
Figure 1 Structure of a PS oligodeoxynucleotide
Figure 2 Cell density versus time in experimental and control groups
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[question] => - Gene fusion, which may produce a hybrid protein or RNA that has altered or abnormal functions compared with the original genes
- Gene alterations, such as mutations or interruptions in a gene
- Disruption of regulatory elements, which can interrupt or relocate elements, such as enhancers or promoters, that control gene expression
- Altered chromatin structure, which can affect the accessibility of genes to transcription factors and RNA polymerase
- Duplications or deletions of certain chromosomal segments, which can increase or decrease the number of copies of specific genes
- Epigenetic changes, such as DNA methylation and histone modifications
In follicular lymphoma, according to the passage, there is excess expression of the bcl-2 protein.
It is unlikely that a frameshift mutation would increase the amount of functional protein available in the cell; this is because frameshift mutations typically result in proteins with altered (usually dysfunctional) sequences. In addition, two of the other remaining choices would decrease bcl-2 expression.
On the other hand, the translocation placing the bcl-2 gene on chromosome 18 near the immunoglobulin heavy chain (IGH) gene promoter on chromosome 14 results in the aberrant activation of the bcl-2 gene by the strong IGH promoter. The IGH promoter is very active in B cells, which produce immunoglobulins, so this explains why the translocation leads to high levels of bcl-2 expression.
A. The translocation moves the bcl-2 gene on chromosome 18 farther from an enhancer region that binds transcription factors and RNA polymerase.
) [1] => Array ( [each_answer] =>B. The translocation results in a frameshift mutation in the bcl-2 gene on chromosome 18, making the protein less susceptible to degradation.
) [2] => Array ( [each_answer] =>C. The translocation places the bcl-2 gene on chromosome 18 near the immunoglobulin heavy chain (IGH) gene promoter on chromosome 14.
) [3] => Array ( [each_answer] =>D. The translocation places the bcl-2 gene on chromosome 18 into the intron of a necessary cellular gene on chromosome 14.
) ) ) [1] => Array ( [quiz_unique_key] => 1403770772 [question] =>This table shows signaling molecules that become dysregulated in certain types of cancer cells. According to the phenotype described in the passage, which changes would most likely be observed in follicular lymphoma B cells?
Reason for Correct Answer:
The passage states that B cells in follicular lymphoma do not divide much faster than normal B cells, but they have an unusually long lifespan.
So, you’re probably not looking for the involvement of a protein that affects the cell cycle.
Even considering this, the choices given in Choices A and B would both inhibit the cell cycle, which would slow cell cycle progression in B cells compared to normal cells, which is very unlikely for cancer cells. The table shows that CDK inhibitors are anti-cell cycle progression, suggesting that CDKs would be pro-cell cycle progression, and their downregulation would block the cell cycle; and, the table also shows that p53 (and its upregulation) is anti-cell cycle progression.
Of the remaining answer choices, downregulation of FLIP expression would be pro-apoptotic, since the FLIP protein is anti-apoptotic.
The B cells in follicular lymphoma, with their long lifespans, experience less apoptosis; this goes with the inhibition of Bax and Bak signaling (which is pro-apoptotic signaling).
) [answers] => Array ( [0] => Array ( [each_answer] =>A. Downregulation of CDKs
) [1] => Array ( [each_answer] =>B. Upregulation of the p53 protein
) [2] => Array ( [each_answer] =>C. Downregulation of FLIP expression
) [3] => Array ( [each_answer] =>D. Inhibition of Bax and Bak signaling
) ) ) [2] => Array ( [quiz_unique_key] => 1403770772 [question] =>What is NOT a likely consequence of early stage follicular lymphoma?
[value] => Array ( [answer] => 4 [description] =>Reason for Correct Answer:
As mentioned in the passage, follicular lymphoma results in B cells clustering in the lymph nodes, so the lymph nodes become enlarged.
The blockage of lymph nodes will reduce the flow of lymphatic fluid, lower the body’s natural immunity and also cause fluid buildup in the regions surrounding the blocked lymph nodes.
Easy bleeding and bruising are not primary symptoms of follicular lymphoma. Despite being a blood disorder, bleeding and bruising would be caused by issues with platelets or clotting factors, not issues with white blood cells.
) [answers] => Array ( [0] => Array ( [each_answer] =>A. Fluid buildup around lymph nodes
) [1] => Array ( [each_answer] =>B. Enlarged lymph nodes
) [2] => Array ( [each_answer] =>C. Lowered immune response
) [3] => Array ( [each_answer] =>D. Easy bleeding and bruising
) ) ) [3] => Array ( [quiz_unique_key] => 1403770772 [question] =>What is a possible explanation for the results of the antisense PS oligonucleotide observed from day 0 to day 2?
[value] => Array ( [answer] => 2 [description] =>Reason for Correct Answer:
The graph shows that the antisense PS oligonucleotide group (the test group) was not significantly different from the sense PS oligonucleotide group (the control group) during the first few days. Notice how the cell density for this group is increasing just like it is in the other groups. However, after this period (marked in red), the cell density begins to decrease. This suggests that the antisense PS oligonucleotide takes some time before it is effective in blocking cell proliferation.
So, you’re looking for an answer that will delay the effect of the PS oligonucleotide.
The replacement of oxygen with sulfur will not significantly affect the charge of the molecule, and, in addition, if the molecule was more reactive, this would not explain the delayed effect. In fact, chemical reactivity is not the mechanism of action of the oligonucleotide. (Choice C is incorrect.)
For a transport protein, a lower Kd does correspond to a stronger interaction and faster rate of transport, but this would hasten the effect of treatment, not delay it. (Choice A is incorrect.)
The passage states that “colonies of B cells” are treated in the experiment. Therefore, delivery to target organs (Choice D) is irrelevant because whole animals are not being used.
Antisense oligonucleotides work by reducing the expression of genes. They do this by binding to mRNA and reducing its translation. Because their mode of action is the reduction in the synthesis of new bcl-2 (rather than the inhibition of bcl-2 itself), it is reasonable to expect a delay in the onset of the phenotypic effect.
) [answers] => Array ( [0] => Array ( [each_answer] =>
A. The Kd value for an oligonucleotide and its transport protein decreases with the addition of sulfur, so initially cells uptake more PS nucleotides.
) [1] => Array ( [each_answer] =>B. Because the mechanism of action is the inhibition of gene expression, there is a delay between delivery and phenotypic outcome.
) [2] => Array ( [each_answer] =>C. The negative charge on the sulfur atom makes the oligonucleotide initially more reactive in intracellular conditions and pH.
) [3] => Array ( [each_answer] =>D. Because PS oligonucleotides have a higher molecular weight, they take longer to reach their target organ(s).
) ) ) [4] => Array ( [quiz_unique_key] => 1403770772 [question] =>The data support which conclusion regarding the effects of the sense oligonucleotide?
I. The sense oligonucleotide decreased the number of cells but not to the extent of the antisense oligonucleotide.
II. The sense oligonucleotide affected the number of cells by binding to bcl-2 mRNA and inhibiting its translation.
III. The sense oligonucleotide temporarily decreased apoptosis.
Reason for Correct Answer:
Look at the results of treatment with the sense (square data points) and antisense (triangular data points) oligonucleotides in Figure 2.
The graph shows that whereas the antisense oligonucleotide decreased the number of cells after day 2, the sense oligonucleotide temporarily increased the number at around day 3. This suggests that the sense oligonucleotide may have temporarily decreased apoptosis/cell death. (Statement III is correct but statement I is not.)
The bcl-2 gene is anti-apoptotic, which gives the follicular lymphoma cells their longer lifespan. Decreasing its mRNA would increase apoptosis, not cell density.
In addition, antisense oligonucleotides are complementary to mRNA, whereas sense oligonucleotides share the mRNA’s code. Inhibition of mRNA translation would be expected for an antisense oligonucleotide, but not for a sense oligonucleotide.
Either point (3) or (4) shows you why statement II is wrong.
A. I only
) [1] => Array ( [each_answer] =>B. II only
) [2] => Array ( [each_answer] =>C. III only
) [3] => Array ( [each_answer] =>D. II and III only
) ) ) [5] => Array ( [quiz_unique_key] => 1325138223 [question] =>Researchers performed an experiment on a cell line not expressing bcl-2 in excess. Cells transfected with a bcl-2 expression vector or a control vector were also treated with the antisense or sense PS oligonucleotide. Which graph shows the expected levels of caspase-3 activity (a marker of apoptosis) after one week?
[value] => Array ( [answer] => 3 [description] =>Reason for Correct Answer:
The question states that the apoptosis level is positively correlated with caspase-3 activity; therefore, higher bars mean more apoptosis.
Because the bcl-2 gene is anti-apoptotic, you would expect lower bars for the bcl-2, if given a treatment that does not block its expression.
According to the passage, the antisense PS oligonucleotide would be expected to reduce the expression level of bcl-2 and thereby negate its effect. Therefore, you would expect higher bars (restored apoptosis) for the antisense oligonucleotide in the bcl-2 conditions.
Because the antisense PS oligonucleotide targets the bcl-2 gene, and the question states that the cell line does express bcl-2 in excess, you would not expect the antisense oligonucleotide to have a significant effect on cells that are transfected with the empty vector.
Choice A matches these expectations.
A. 
B. 
C. 
D. 