Submit reasoning and process, not merely answers! 3.1. A child presents to you,
ID: 3506954 • Letter: S
Question
Submit reasoning and process, not merely answers! 3.1. A child presents to you, her pediatrician, with all the classical symptoms of diabetes. Upon testing, you find that antibody against insulin detects very low levels of insulin in her blood, but she responds normally to administered insulin. You are surprised to find, however, that the same antibody detects levels of insulin in the pancreas that are grossly higher than normal. What mutation might account for these findings? 3.2. An enzyme has a molecular weight of 60,000 daltons. When it is exposed to detergent, the protein breaks up to identical inactive components with molecular weights of 20,000 daltons. If the detergent is removed by dialysis, the 60,000-dalton protein reforms and regains enzymatic activity You have isolated two mutant proteins. Mutant I shows no enzymatic activity and has a molecular weight of 20,000 daltons whether or not detergent is present. Mutant 2 has a molecular weight of 60,000 without detergent and 20,000 with detergent but shows no enzymatic activity in either case. a. Suggest defects to explain the behavior of each of the mutant enzymes. b. A person is heterozygous for Mutant 2 (i.e., has 50% Mutant 2 polypeptide and 50% normal polypeptide). How would you explain an observation that the person has 87.5% of the enzyrnatic activity of a normal person? How would you explain an observation of 12.5% activity Ascribe the terms "dominant" or "recessive" to the mutation leading to Mutant 2, according to the two situations presented in b. c.Explanation / Answer
3.1. The mutation must be affecting the proper folding of the protein. So, an improperly folded protein is unstable and cannot perform its function. Hence the antibody detects low levels of insulin in the blood. The same antibody could detect high levels of insulin in the pancreas because the unfolded protein is non-functional. So, the pancreas produces more protein. However, it gets degraded once it enters the bloodstream.
3.2
Native protein MW = 60 kDa
Molecular composition = Trimer (With each subunit MW = 20 kDa)
Mutant I: Dialysis cannot restore the enzymatic activity. So, the mutation affects non-covalent interactions. The mutation must be in one of the interacting surfaces that provide strong association by ionic/polar interactions.
Mutant II: The mutation must be located in the active/substrate binding site. So, the presence or absence of the detergent has no difference in the enzyme activity. It is constitutively inactive.
If the mutation is dominant, only 12.5% activity is observed.
If the mutation is recessive, 87.5% activity is observed.