Blood type O means the blood cells do NOT express antigen A and antigen B, which means the individual will be antibody A and antibody B, he would die if we transfuse the individual with blood type A and blood type B because the antibody A and B in that individual will attack the transfused blood.
Therefore the answer is Blood Type O.
Halohydrin are alcohol with a hydrogen replaced with a halogen (the hydroxy group and the halogen are adjacent). The halohydrin dehydrogenase will convert that into an epoxide, getting rid of the halogen and a hydrogen, they will form hydrohalic acid and therefore free proton in the aqueous solution. Therefore, the answer is:
When the enzyme activity of halohydrin dehydrogenase increase, more protons ($ H^+ $) are released in the catalyzed reaction.
More protons means lower pH means more yellow, therefore the answer is Mutant D.
There is no way I can answer this question from the question alone. It requires context from the lecture video. The goal of the ISM is to reduce oxygen activity and make sure the electrons are get from the electrodes when one want to use the engineered enzyme to measure blood glucose level , therefore the answer is Mutant B because that mutant leads to reduced oxygen activity.
We want the enzyme to use NADH, therefore, we are looking for increased NADH affinity and increased catalytic rate.
I failed this one and I still do not understand, the $ K_m $ for NADH will obviously increase because the equilbrium would shift towards forming the enzyme substrate complex, what I don't understand is why will have lower $ K_{cat} $, doesn't that means the reaction shift towards not creating the product?
This is basically the same as question 3 above, increased activity, decreased pH, increase in yellow signal.
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