BCH210 RUNBACK EXAM QUESTIONS WITH ANSWERS GRADED A+
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Module
BCH210 RUNBACK
Institution
BCH210 RUNBACK
BCH210 RUNBACK EXAM QUESTIONS WITH ANSWERS GRADED A+
Lactate dehydrogenase - Answer-Pyruvate + NADH --> lactate + NAD+
Regenerates NAD+ for glycolysis when O2 is unavailable; in red blood cells, lactate is always produced since no mitochondria
Cori cycle - Answer-1. Lactate is produced ...
Regenerates NAD+ for glycolysis when O2 is unavailable; in red blood cells, lactate is
always produced since no mitochondria
Cori cycle - Answer-1. Lactate is produced in the bloodstream by red blood cells
2. Liver picks up lactate; lactate dehydrogenase converts it back into pyruvate
3. Pyruvate is converted into oxaloacetate, where it either enters the CAC or GNG
Gluconeogenesis overview - Answer-Occurs in liver/kidney to produce glucose; only the
conversion of pyruvate to oxaloacetate occurs in the mitochondria
Only the reversible steps are shared with glycolysis; 4 enzymes are used to catalyze a
pathway with delta G = -38
Pyruvate carboxylase & PEP carboxykinase - Answer-Reverse of glycolysis step 10
(making PEP):
1) pyruvate + bicarbonate + ATP --> oxaloacetate + ADP + P
2) oxaloacetate + GTP --> PEP + GDP + CO2
Bicarbonate donates CO2 to pyruvate; CO2 release is then used to fuel oxaloacetate
phosphorylation to PEP
Gluconeogenesis pathway - Answer-1. Pyruvate is converted to oxaloacetate, then PEP
2. 2 PEPs are turned into GAP and DHAP --> F1,6BP
3. F1,6BP has a phosphate cut off to F6P
4. F6P is converted back to G6P
5. G6P is returned to glucose
Tandem enzyme (PFK-2/fructose-2,6-bisphosphatase) - Answer-F2,6BP stimulates
PFK-1 of glycolysis and inhibits fructose-1,6-bisphosphatase of GNG
Tandem enzyme of PFK-2 and fructose-2,6-bisphosphatase controls F2,6BP; single
polypeptide chain with 3 domains -- 2 active sites for the enzymes and a regulatory
domain
Phosphorylation of the regulatory domain determines which enzyme is active
Low blood glucose: PKA phosphorylates the domain, promoting gluconeogenesis so the
liver will share glucose
High F6P levels: phosphoprotein phosphatase dephosphorylates the domain, promoting
glycolysis
Triacylglycerol - Answer-Storage form for lipids; glycerol backbone connected to three
fatty acyl chains via ester linkages
Triacylglycerol lipase (hormone sensitive lipase) - Answer-Removes two tails from
triacylglycerol; activated by PKA
Monoacylglycerol removes the third tail
Albumin - Answer-Transporter for released acyl chains (amphipathic)
Fatty acid activation - Answer-Upon arrival in a cell, fatty acids are activated by adding
CoA so they can't leave the cytoplasm
Overall, requires 2 ATP and 1 H2O
Acyl CoA synthetase - Answer-Fatty acid activation:
fatty acid + ATP + CoA --> fatty acid-CoA + AMP + PP
This reaction is driven forward by the subsequent hydrolysis of pyrophosphate, which
requires H2O
AMP inefficiency - Answer-Any reaction that produces AMP requires an extra ATP to
remake ADP; adenylate kinase catalyzes ATP + AMP --> 2 ADP
Carnitine shuttle - Answer-Since beta-oxidation occurs in the matrix, but CoA can't cross
the inner membrane, an intermembrane shuttle is needed:
1. CAT-1 transfers acyl from CoA to carnitine
2. Translocase swaps intermembrane acyl-carnitine for matrix carnitine
3. CAT-2 transfers acyl back to CoA
, Beta-oxidation - Answer-Each round, H2O/CoA are used to shorten the acyl chain by 2
carbons, releasing acetyl CoA/NADH/FADH2
1. Acyl CoA dehydrogenase: creates a double-bond between alpha-beta carbons,
releasing FADH2
2. Hydratase: adds H2O across the double bond
3. Dehydrogenase: creates a keto group on the beta carbon, releasing NADH
4. Thiolase: new CoA releases acetyl CoA from the beta-keto
Monosaccharides - Answer-Soluble in water due to OH groups (always have either an
aldehyde or ketone)
Formula: (CH2O)n, where n is at least 3
Carbohydrate numbering - Answer-Always starts at the carbonyl; aldehyde is C1,
ketone is C2
L vs. D isomers - Answer-Named based on left (l) vs. right (d) orientation of OH group
Capital lettering given to highest-number chiral carbon; D is physiologically relevant
Number of stereoisomers - Answer-2^x, where x is the number of chiral carbons
x = total number of carbons - 2 (aldehyde) or -3 (ketone)
Carbohydrate cyclization - Answer-Hydroxyl group initiates a nucleophilic attack on
same-chain carbonyl carbon, donating an electron pair
Must occur in solution as water donates/receives protons
Easily reversible
Furan - Answer-5-member ring (4C 1O)
Pyran - Answer-6-member ring (5C 1O); formed when attacking hydroxyl group is
further down in the sugar structure
Hemiacetal - Answer-Aldehyde derivative
Hemiketal - Answer-Ketone derivative
Glucopyranose formation - Answer-Glucose's C5 hydroxyl attacks the aldehyde, forming
a 6-member ring with C1-C5 and the attacking hydroxyl's O; if a C2 ketone was
attacked, C2-C5 would form a 5-member ring
Alpha vs. beta glucopyranose - Answer-Alpha: OH points below ring
Beta: OH points above ring (more stable due to less repulsion between OH groups)
Refers to the OH (C1) formed when carbonyl O picks up an H+ from water, not the in-
ring O from the attacking hydroxyl
Honey - Answer-Fructose is the main sugar found in honey
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