🧬AP Biology
🧬AP Biology
FRQ 1 – Interpreting and Evaluating Experimental Results (Long)
Unit 1: Chemistry of Life
FRQ 1 – Interpreting and Evaluating Experimental Results (Long)
Unit 1: Chemistry of Life
Practice FRQ 1 of 201/20
1. Proteins are essential macromolecules that perform a diverse range of functions in living organisms, including catalyzing metabolic reactions. The specific function of a protein is determined by its three-dimensional structure.
Pectinases are enzymes that break down pectin, a polysaccharide found in plant cell walls. Researchers isolated a wild-type strain of the fungus Aspergillus niger that produces pectinase and a mutant strain that produces a modified form of the enzyme. To compare the catalytic efficiency of the two enzymes, the researchers added equal amounts of purified wild-type or mutant pectinase to separate solutions containing 100 mg of pectin. A third solution containing pectin but no enzyme served as a control.
The reaction mixtures were incubated at 37°C. Every 10 minutes for 40 minutes, the researchers measured the mass of pectin remaining in each mixture. The results are presented in Figure 1.
To investigate the structural basis for the difference in enzyme activity, the researchers sequenced the gene encoding the pectinase enzyme in both strains. They identified a single nucleotide substitution in the mutant gene that resulted in an amino acid change. Based on the sequence data, the researchers constructed a structural model of the active site region for both the wild-type and mutant enzymes (Figure 2). They also performed a second experiment to test the thermal stability of the enzymes by incubating them at a range of temperatures from 30°C to 80°C and measuring the percent of enzyme activity remaining.
A. Describe how the R-groups of amino acids contribute to the tertiary structure of a protein.
Figure 1. Mass of pectin remaining (mg) over time (minutes) at 37°C in reaction mixtures containing 100 mg pectin plus either wild-type pectinase, mutant pectinase, or no enzyme (control). Error bars show ±2SE at each time point.
B.
i. Identify the dependent variable in the experiment shown in Figure 1.
ii. Justify the inclusion of the reaction mixture containing no enzyme in the experiment.
iii. Based on Figure 1, describe the difference in the rate of pectin breakdown between the wild-type and mutant enzymes during the first 20 minutes.
Figure 2. Structural model of the same loop region in wild-type and mutant pectinase. Wild-type contains a disulfide bond between two cysteine side chains; mutant has a serine substitution at one position and therefore lacks the disulfide bond.
C.
i. Identify the independent variable in the thermal stability experiment described in the transition.
ii. Based on Figure 2, identify the specific type of chemical bond present in the wild-type enzyme that is absent in the mutant enzyme.
iii. The gene segment encoding the loop region shown in Figure 2 is composed of 120 nucleotides. Calculate the number of amino acids in this loop region.
D.
i. The researchers claim that the mutant enzyme is less efficient at breaking down pectin than the wild-type enzyme. Using data from Figure 1, support the researchers' claim.
ii. The researchers predict that the mutant enzyme will lose its activity more rapidly than the wild-type enzyme when incubated at 60°C. Justify this prediction based on the structural difference shown in Figure 2.
