🧬AP Biology
🧬AP Biology
FRQ 1 – Interpreting and Evaluating Experimental Results (Long)
Unit 1: Chemistry of Life
Practice FRQ 1 of 201/20
1. Proteins, such as enzymes, rely on their specific three-dimensional structures to function correctly within biological systems. The stability of this structure is maintained by various chemical interactions between the amino acids that make up the protein chain.
Researchers are investigating the structural stability of Enzyme Z, a digestive enzyme found in the stomach of a certain mammal. To understand the importance of specific amino acids in the enzyme's structure, the researchers created a mutant strain of the enzyme. In the mutant enzyme, a positively charged Arginine amino acid at position 105 was replaced with a negatively charged Aspartic Acid. The researchers hypothesized that this mutation would affect the enzyme's stability and function.
To test this hypothesis, the researchers incubated both the Wild Type and Mutant enzymes at three different pH levels: pH 5, pH 7, and pH 9. After incubation, they measured the relative activity of each enzyme form. The results were compared to the maximum activity observed in the Wild Type enzyme at pH 7 (Figure 1).
To investigate the molecular mechanism responsible for the observed differences in enzyme activity, the researchers used computer modeling to visualize the protein structure. They focused specifically on the interaction between the amino acid at position 105 and a Glutamic Acid residue located at position 200 in the folded protein. The structural models for both the Wild Type and Mutant enzymes are shown in Figure 2.
A. Describe how the R-groups of amino acids contribute to the tertiary structure of a protein.
Figure 1. Relative activity of Wild Type and Mutant Enzyme Z after incubation at pH 5, pH 7, and pH 9. All bar heights are percent activity relative to the Wild Type enzyme at pH 7, defined as 100%. Error bars show ±SE (±5 percentage points) for every bar.
B.
i. Identify the dependent variable in the experiment shown in Figure 1.
ii. Justify the researchers' decision to measure the activity of the Wild Type enzyme at pH 5 and pH 9 in addition to pH 7.
iii. Based on Figure 1, describe the effect of the mutation on the relative activity of Enzyme Z at pH 7.
Figure 2. Tertiary-structure interaction between amino acid 105 and amino acid 200 in Enzyme Z. Panel A (Wild Type) shows a stabilizing ionic bond between Arg105 (positive) and Glu200 (negative). Panel B (Mutant) shows electrostatic repulsion between Asp105 (negative) and Glu200 (negative).
C.
i. Based on Figure 2, identify the specific type of chemical bond that forms between the amino acids in the Wild Type protein.
ii. Based on Figure 2, identify the specific characteristic of the amino acid R-groups in the Mutant protein that prevents the formation of the stabilizing bond observed in the Wild Type.
iii. The mRNA sequence encoding the Wild Type enzyme is 960 nucleotides long (including the stop codon). Assuming each nucleotide triplet encodes one amino acid, calculate the number of amino acids in the functional protein (excluding the stop codon).
D.
i. Researchers claim that the Mutant enzyme is less efficient than the Wild Type enzyme because the mutation disrupts the protein's tertiary structure. Using data from Figure 1, support the researchers' claim.
ii. Justify the researchers' claim by explaining how the interaction shown in Figure 2 for the Mutant protein affects the enzyme's active site.
