Biochemistry Practice Passage Explanations

9.3 Getting the Edge in Biochemistry

Getting a high score on Test Day requires a very solid foundation in biochemistry because this subject area is heavily tested. However, it is important to develop a grasp of biochemistry in the context of the other sciences, especially biology, organic chemistry, and general chemistry. Many of the passages you encounter on Test Day require you to make connections between the other sciences and biochemistry. In addition, biochemical processes do not occur in isolation. Many of the passages and questions focus on how changes at the molecular level result in changes in the entire organism’s physiology.

On Test Day, biochemistry passages will feature a variety of figures, including graphs, charts, molecular structures, chemical equations, and visual representations of biochemical concepts. Be prepared to analyze a variety of different images because you will be repeatedly asked to do so on Test Day.

The key to this question is a careful reading of the question stem and the passage. The question stem asks us which plate will have the highest and lowest percentages of partially translated proteins. Paragraph 2 tells us that telithromycin is more potent than erythromycin, so we might be led to predict that the order is plate 2 (telithromycin) > plate 1 (erythromycin) > plate 3 (control). That would lead us to (D), which is a trap. Paragraph 3 tells us that although telithromycin is more potent as an antibiotic, it is actually worse at inhibiting synthesis than erythromycin is! So the correct answer is actually plate 1 > plate 2 > plate 3, which matches (C).

Paragraph 3 states that the mRNA sequence shown in Figure 2 makes a peptide macrolide resistant; it can escape the tunnel even if erythromycin or telithromycin is bound. If we want the greatest increase in inhibition, we need a significant change in the sequence of that peptide. To answer this question, we’ll need to use the genetic code in Figure 2 and look for a mutation that would lead to a different amino acid. By cycling through the choices, we find that the correct answer is (D): AAC codes for asparagine, whereas AGC codes for serine.

In (A), proximity to the start codon does not, by itself, increase the likelihood that a mutation would affect inhibition by macrolides. In (B), the third nucleotide in the codon is in the wobble position, which tends to be the most likely nucleotide to result in a silent mutation when altered. (C) might look promising: it changes the first nucleotide in the codon. A look at the genetic code, though, shows us that this is one of the few cases where such a mutation is actually silent; both CUG and UUG code for leucine.

We’re looking for a change likely to happen in the presence of erythromycin. According to paragraph 4, erythromycin can cause the nascent peptide to fall out of the ribosome. However, because this happens without a stop codon, the nascent peptide is not released from the tRNA. So we would expect a buildup of peptidyl-tRNA, or peptides still bound to tRNA molecules. This matches (D).

In (A), because protein synthesis decreases, we would not expect the concentration of charged tRNAs to drop; if anything, they would increase. In (B), this would be true if erythromycin bound the two ribosomal subunits together, but nowhere does the passage imply this happens. In (C), paragraph 4 states that this kind of mutation results in the expression of proteins that would normally be stopped by erythromycin. However, it is unlikely that such a mutation would happen spontaneously in the first few minutes after erythromycin administration.

The graph in the question stem represents the products of protein synthesis in normal E. coli cells. The graphs in the answer choices represent possible protein synthesis in E. coli after exposure to telithromycin. According to paragraph 3, the inhibition of synthesis is determined only by the nature of the N-terminal sequence of the peptide. Because we have no reason to believe that there is a specific correlation between N-terminal sequence and protein size, we would expect an overall decrease in protein synthesis at all molecular masses. This matches (B).

(A) would be correct if the inhibition was based solely on molecular size (in this case, inhibiting synthesis of proteins >150 kDa in mass). In (C), the total amount of protein produced is the same; according to paragraph 4, protein synthesis drops when macrolides are given. (D) shows partial inhibition, as the passage states, but only for proteins with high molecular weights; as with (A), we have no reason to believe the inhibition is limited to certain molecular weights.

Where would we find radiolabeled methionine after five minutes? Every peptide chain begins with methionine; the only codon for Met also is the start codon. However, the majority of finished peptides do not have Met at their N-terminus. In most proteins, that initial Met, along with other N-terminal residues, are removed in post-translational processing. Among other things, that N-terminus sequence is often involved in signaling the destination of a protein (for example, whether it should end up in the cell membrane). There also is no mechanism for exchanging methionine in existing proteins, so the correct answer is (B).

In (A), most initial methionines are lost in post-translational processing. For (C), most proteins have internal methionines, but most of the initial methionines are removed. There is no mechanism in (D) for exchanging amino acids in existing proteins with new amino acids.

🎯 Biochemistry Practice Passage Explanations

9.3 Getting the Edge in Biochemistry