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ANSWER (1a): 4.8mg lactose/100mL means [lactose] = 1.4 x 10- (^4) M or .14 mM. As Km = 0.9mM, [S] is the same order-of-magnitude as Km, but slightly below it. ANSWER (1b): For 1/2 – maximal velocity, [S] = Km. We have [S] < Km, so the rate is < Vmax/2. ANSWER (1c): Vo = (^) KVm max +^ [ [ SS ]] = V .9max +^ • .14^ .14 = .13 V max i.e., the velocity will be 13% of the maximum. ANSWER (1d): Remember that Vmax depends on [E]total. For 200 mg lactase: V Vmaxmax = (3.5= 700μ μM/minM/min/mg)(200mg) From ( c ), vo = .13Vmax = .13 (700 μM/min) = 91 μM/min From (a), the [lactose] in milk is 0.14 mM, so ½ is 0.07 mM (or 70 μM). So, at a rate (vo) of 91 μM/min, it will take 70 μM/91 μM/min = 0.77 min = 46 s. ANSWER (2):
AND^ So, ratio of velocities for S^1 & S^2 = 9. velocity measured in Tube 2 velocity measured in Tube 1 = 9. So Tube 2 had S 1 in it; Tube 1 had S 2. To calculate Vmax, just substitute all the numbers.
Vo(S 1 ) Vo (S 2 ) =^ [S 1 ](K Sm^2 + [S 2 ]) [S 2 ](Km^ S^1 + [S 1 ]) = K Km^ Sm S^21 ++^ [S[S 12 ]] = (^) 2.0mM^20 mM^ ++^ 0.1mM 0.1mM
(because [S^ (because Vmax 1 ] = [S^ same) 2 ]) (substituting numbers)
Vo (S 1 ) = (^) KV Smmax (^1) +^ [S [S^1 1 ]] & Vo (S 2 ) = (^) KVm Smax (^2) +^ [S [S^2 2 ]]
So, Vmax = 100.8 ≅ 101 μmol/min (or 100.5 if you use the numbers for S 2 – rounding errors.) ANSWER (3a): Glu [A-] / [HA] = 0.2, i.e. 5 times as much HA as A-35 w/pKa=5.9 will be predominantly in the protonated (neutral) form (by Henderson- (^) ) -Hasselbalch,
Asp-52 w/pKa = 4.5 will be predominantly in the ionized, charged state, with 5 times as much A-^ as HA. ANSWER (3b): Hypothesis: Glu (i.e. proton acceptor) or may help to stabilize a transient-35 acts as an acid (i.e. proton donor) in the mechanism. Asply charged species that occurs along the reaction-52 may act as either a base pathway. ANSWER (3c): Mutant to test whether Glu35 serves as acid would be Glu35Gln (E35Q). [This mutant has been made and tested w/results that it has virtually no catalytic activity (0.1% of wild-type) but binds substrate as well as wild formation of ES.]-type. This shows that the side-chain of Glu35 serves as a proton donor but is not involved in
Mutants to test whether neg. charge on Asp catalytic activity but binds substrate as well as wild-52 is important : Asp52Asn [This mutant also has n – type. This implicates the neg. charge of Aspo-52 as important in catalysis, not formation of ES.] Double mutants first. In this case, since each of the single mutants are essentially dead, there wouldn’t be much-mutants : in general, these are more difficult to interpret. It’s usually best to make single-site too learn from the double mutant. A possible double mutant : Glu35Asp/Asp52Glu might give some insight about steric constraints in the active site.
ANSWER (4a): hexokinase D will close to its maximal rate because [S]>>K its 1/2-maximal rate because [S]< Km. m. In contrast, glucokinase will work at under
ANSWER (4b): As hexokinase D is already on the flat, asymptoti to an increase in glucose levels. Glucokinase is on the steep part of the curve and will therefore be quitec part of its Vo vs [S] curve, it will not be very responsive responsive. That is, its velocity will increase significantly with an increase in blood [glucose].
Vo (S 1 ) = 4.8 = (^) 2.0VmaxmM^ •^ +0.1 0.1mMmM
If soybean trypsin inhibitor is not removed it can inhibit the t nutritional benefits of the Tofu (its proteins will not be effectively cleared) and may lead to intestinalrypsin in your intestine. This can lower the upset. ANSWER (7): Lys and Arg residues, being hydrophilic and charged, are apt to be fou including trypsin. So, a trypsin molecule will find lots of potential substrates (i.e., Lys and/or Arg) tond on the surface of proteins, cleave on other trypsin molecules (i.e., autolysis.) In contrast, the substrates for chymotrypsin (large hydrophobic residues) will be mainly buried on the interior of chymotypsin, making it a worse substrate for itself.
