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Anyone help me in these doubts !
Nov 2013/p11 ----->>
Q11....Detailed explanation pls !
- Q6 ...why is it C ,when the first carbon atom of the two B-glucose doesn't contain the OH group up?
- B may be correct but yet there is no gylcosidic bond formed ? soo ?
Q15 ....I know the water will enter the cell but ....wht is abt the inner and thicker walls ?
Q19...why B ?
Q26 .... ??
6- Monomers are B glucose & in order for them to form glycosidic bonds successive B glucose molecules have to be flipped to 180 degrees i.e upside down. ( If they are not flipped glycosidic bond cannot be formed so cellulose is not formed)
11-Competitive inhibitor binds the enzyme where the substrate is supposed to bind i.e the active site. If the conc of substrate is increased, the chance of the substrate binding with the active site before the inhibitor is higher. (Non competitive inhibitor binds the allosteric site & increasing the conc of substrate doesn't reduce their effect)
so statement 1 & 2 are correct
statement-3 is Incorrect as enzymes reduce the activation energy not inhibitors.
statement-4 is Incorrect as non competitive inhibitors reduce the max rate of reaction not competitive inhibitors.
15-Not sure about this one but I guess the thick outer walls are lignified so, it won't allow water to enter.
19- Total 19 base pairs,( complementary base pairing A with T & C with G)
strand 1 has 4-T so strand 2 will have 4-A
strand 2 has 5-T so strand 1 will have 5-A
strand 2 has 7-C so strand 1 will have 7-G
base pairs = 4+5+7=16 so 19 - 16=3
26-What exactly do you have problem with here?