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- The plasma profiles of codeine (COD) and metabolites for 2 individuals (labeled A and B) are shown below. The X-axis is time in hours after an oral dose of codeine. [M=morphine; C6G=COD-6-glucuronide; M3G = morphine-3-glucuronide; NM (ignore)]. Note the data is shown on a log scale on the Y-axis. (A) Which individual is the poor metabolizer? Explain how you know this from the profiles? (B) Is this a problem for cough suppression? Explain. -CH HO Codeine COD 10 000 1000 C6G COD 100 M3G M6G NM 10 M 10 20 30 0 10 20 30 Plasma concentration (nmol I-)Serum blood of a patient with dislipoproteinemia type 1 has milky appearance even in fasting. If serum stays at low temperature (40) for several hours fatty layer appears on its surface. What are the possible causes of these symptoms? To explain this, answer the questions and do the following tasks: a) what compounds of serum must be tested for that patient in biochemical lab? b) write the reaction which does not occur properly in patient’s blood; c) write down the schemes, explaining how the products of the previous reaction are used in adipose tissue and heart in healthy person 2 hours after a meal.1) Draw a chromatogram depicting separation of proteins 1, 2, and 3 (protein #1, mw 30,000, protein #2, mw 60,000, and protein #3, mw 90,000). b)Indicate on the chromatogram in (a) where insulin would elute. Would insulin have total accessibility, some accessibility or no accessibility to the particle pores on this column? c)Referring to the size exclusion chromatogram in (a), indicate on the chromatogram where you would expect a molecule with a size of 40,000 to elute. Would you expect the molecule to be well resolved, somewhat resolved, or not resolved from the peak for protein #1?
- Hemoglobin glycation (so named to distinguish it from glycosylation, which is the enzymatic transfer of glucose to a protein) is a non-enzymatic process that involves reaction of the N-terminal amino group of hemoglobin and glucose. The amount of glycated hemoglobin (GHB) is usually about 5% of total hemoglobin (and corresponds to a blood glucose concentration of 120 mg/100 mL). However, in people with untreated diabetes this value may be as high as 13%, which indicates an average blood level of about 300 mg/100 mL -dangerously high. One of the aims of insulin therapy is to maintain GHB values of about 7%. Draw a possible chemical scheme for the glycation of hemoglobin.From the three figures below, indicate which represents best the catalytic pocket of trypsin, a serine protease, secreted by acinar pancreatic cells. C) A) A. Figure A. B.Figure B. ⒸC. Figure C. Asp 189 B) Asp 189 Val 190 Val 216 work of Val 216 Val 190Penicillin is hydrolyzed and thereby rendered inactive by penicillinase, an enzyme present in some resistant bacteria. The mass of this enzyme in Staphylococcus aureus is 29.6 kd. The amount of penicillin hydrolyzed in 1 minute in 10 ml solution containing 10–9 g of purified penicillinase was measured as a function of the concentration of penicillin. Assume that the concentration of penicillin does not change during the assay. [Penicillin] µM Amount hydrolyzed (nanomoles) 1 0.11 3 0.25 5 0.34 10 0.45 30 0.58 50 0.61 a) Plot V0 versus [S] and 1/V0 1/[S]. Does penicillinase appear to obey Michaelis-Menten kinetics? b) What is the value of KM? c) What is the value of Vmax? d) What is the turnover number of penicillinase under these conditions?
- Homocystinuria is caused by a defect in cystathionine beta-synthase (or 13-synthase), which leads to an accumulation of homocysteine in the blood. This accumulation causes symptoms such as a tall, thin frame, flushed cheeks, and osteoporosis (thinning of the bones). These individuals should limit their intake of proteins that contain methionine, such as egg whites. Using your understanding of biochemistry, explain why people with Homocystinuria should not consume egg whites and other such proteins.In germ-free mice, which harbor no intestinal bacteria, the O-linked oligosaccharides of intestinal glycoproteins tend to lack a terminal fucose residue (see Fig.). (a) What enzyme is not produced in normal quantities in these mice? (b) What monosaccharides tend to appear at the ends of O-linked oligosaccharides in these animals?Hemoglobin molecules exposed to high levels of glucose areconverted to glycated products. The most common, referred to as hemoglobin A1C (HbA1C), contains a b-chain glycatedadduct. Because red blood cells last about 3 months, HbA1Cconcentration is a useful measure of a patient’s blood sugarcontrol. In general terms, describe why and how HbA1Cforms.
- Sketch the appearance after visualization of a protein mixture containing the seven proteins(ovalbumin, insulin, fibrinogen, y-globulin, collagen, hemoglobin, myoglobin when subjected to SDS-PAGE.The intravenous infusion of fructose into healthy volunteers leads to a two- to fivefold increase in the level of lactate in the blood, a far greater increase than that observed after the infusion of the same amount of glucose. (a) Why is glycolysis more rapid after the infusion of fructose? (b) Fructose has been used in place of glucose for intravenous feeding. Why is this use of fructose unwise?(b)( ) In mature erythrocytes (red blood cells) the end product of glycolysis is lactate because of the absence of mitochondria. On the right is a table comparing the rate of lac- tate production in hemolysates (lysed cells) of human RBCs as a function of pH with dif- ferent substrates introduced into the glyco- lytic pathway. The hemolysate was fortified with 30 μmoles substrate, 7.5 μmoles MgCl2, 10 μmoles disodium phosphate, 1.5 μmoles NAD* and 5 μmoles ATP in a volume of 5 TABLE 3-LACTATE PRODUCTION IN FORTIFIED HEMOLYSATES OF HUMAN ERYTHROCYTES* Substrate Glucose Glucose Glucose-6-phosphate Glucose-6-phosphate Fructose-1,6-diphosphate Fructose-1,6-diphosphate Lactate production† No. of experiments pH 6 7.1 2.03 ± 0.91 6 7.8 4.76 ± 1.09 5 7.1 10-731-88 5 7.8 12.34 ±2.92 5 7.0 7.15±0.73 5 7.7 7.15±0.80 mL. The rate of lactate production is given as μmoles of lactate/g Hb/hr at 37° C, buffered to either pH 7.1 or 7.8, as indicated. According to the results in the table which…