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2 years ago

Which of the following statements best helps explain the reaction specificity of an enzyme?

Biology

1 answer:

irakobra [83]2 years ago

8 0

Answer:

This question lacks options, the options are:

A) The free energy of the reactants is greater than the free energy of the products.

B) The equilibrium constant of the reaction is much greater than 1.

C) The shape and charge of the substrates are compatible with the active site of the enzyme.

D) The concentration of the enzyme inside living cells is greater than the concentration of substrate.

The answer is C

Explanation:

An enzyme is a proteinous molecule that increases the rate of reaction of a biochemical reaction in living organisms. However, a protein molecule contains amino acids which makes up it's primary structure. An enzyme performs it action by binding to it's substrate using it's active site.

An enzyme is substrate specific because the shape and charge of the substrates are compatible with the active site of the enzyme, where the reaction occurs. This means that an enzyme will not bind to any substrate that its active site does not recognize

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Two autosomal genes, J and K, are 60 map units apart. You perform the following testcross: J K / j k x j k / j k. At what freque

Dmitry [639]

Answer:

J K / j k = 20%

j k / j k = 20%

J k / j k = 30%

j K / j k = 30%

Explanation:

To calculate the recombination frequency, we have to know that 1% of recombinations = 1 map unit = 1cm. And that the maximum recombination frequency is always 50%.

The map unit is the distance between the pair of genes for which every 100 meiotic products, one of them results in a recombinant one.

So, en the exposed example:

J and K are autosomal genes

J and K are separated by 60 M.U.

60 M.U. means that there is 60% of recombination.

Cross) J K / j k x j k / j k

Gametes) JK Parental jk, jk, jk, jk

jk Parental

Jk Recombinant

jK Recombinant

One map unit equals 1% of recombination frequency. This means that every 100 meiotic products, one of them is a recombinant one.

1 M.U. -------------- 1% recombination

60 M.U. ------------ 60% recombination

30% Jk + 30% jK

100 M.U. - 60 M.U. = 40 M.U.

40M.U.--------------40 % Parental (Not recombinant)

20% JK + 20% jk

Punnet Square) JK jk Jk jK

jk JK/jk jk/jk Jk/jk jK/jk

J K / j k = 20%

j k / j k = 20%

J k / j k = 30%

j K / j k = 30%

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