5. Phenylketonuria (PKU) is an inherited disease which results from the lack of

Question

5. Phenylketonuria (PKU) is an inherited disease which results from the lack of the enzyme phenylalanine hydroxylase (PAH). The PAH enzyme catalyzes the first step in the degradation of Phe. In PKU patients, Phe accumulates and is eventually transaminated to phenylpyruvate. Excess phenylpyruvate accumulates in the blood and urine and has the effect of causing mental retardation if untreated. Screening programs identify PKU babies at birth, and treatment consists of a low Phe diet until maturation of the brain is completed. It is known that the PAH enzyme contains 451 residues and has a MW of 51.9 kDa. More than 60 different mutant genes giving rise to nonfunctional PAH proteins have been identified in PKU patients. a. In order to learn more about PAH, 3 isozymes were isolated from rat liver. Their MW ace. identical, but the pl values are 5.0, 5.6 and 6.1. A DEAE anion exchange cellulose column was used to separate the isozymes. What order of elution would you expect the isozymes to elute if a buffer at pH 5.8 is used along with a gradient of NaCI from 100 mM to 500 mM over 10 minutes? I would expect the elution to be 6.1,5.6, and lastly 5.0 b. Once purified, the investigators wanted to determine if Phe served as a regulator of PAH beyond being a substrate. Polyacrylamide gel (under denaturing and non-denaturing conditions) electrophoresis was performed and the results are shown in Fig. 1 below. Explain these results and your reasoning. (limit 200 words) (2 points) C. Next, kinetic studies were performed on PAH. A plot of velocity Phe concentration yielded a sigmoidally shaped curve. Explain these results and your reasoning. d. More kinetic data is shown in Fig. 2. interpret the data based on relative activity The effects of the hormones glucagon and insulin on PAH activity are shown in Fig. 3 below. Explain these results and your reasoning e. f. Tyr is not an essential amino acid in normal people, but it is essential in PKU patients. Give an explanation for this.

Explanation / Answer

For B. Remember that the absence or presence of SDS is important for electrophoresis, in the first column having SDS the protein is denatured which means that it is dissociated in each of its subunits, so it will have greater mobility and appear to weigh less molecular than what it actually has or if it were native. in the second column it marks that it does not have SDS nor Phe, for which the enzyme is alone, and has a certain spatial conformation each of its subunits, the last column has its substrate (Phe) an enzyme to be in the presence of its substrate changes the spatial conformation of the whole protein, although the active site is very small compared to the whole enzyme, the reaction will end up moving all the subunits.

For C: Most of the enzymes behave in a rectangular hyperbole manner (that is, they do not follow the Michaelis-Menten kinetics), however there are enzymes that are distinguished by having another type of regulation known as alosteric enzymes which are distinguished by their response to changes in substrate concentration. The curves of these enzymes are called sigmoid because they are S-shaped. Most alosteric enzymes show sigmoid kinetics, which are given by the existence of cooperativity among their subunits.

For D: It is obvious that you give this result if there is no preincubacion the substrata to be colder do not interact enough with the enzymes. This obeys the "law of mass action" that is why it must have preincubacion so that both the enzyme as the substrate and the medium are at the same temperature and the reactions are optimal.

For E: The first column is our negative control, evidence that there can be no activity without the substrate. In the last column we find a basal level of activity given that the enzyme is with its substrate (works as a positive control). In the second column is the same although there is insulin there is no substrate and therefore there can be no activity. The third column should behave like the second column, however, it is not like that, and in the fourth column it is where we have the most activity. One might think that the terminal amino acid of glucagon would have to be Phe or failing that there must be Phe amino acids exposed in its structure which are being degraded by the enzyme (which would justify the result of the fourth column in which besides to be acting on the Phe also does it on the glucagon).

For F: The precursor of tyrosine is Phe, patients not having a good metabolic pool of Phe to synthesize the tyrosine they need require tyrosine.

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