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lesya692 [45]
3 years ago
15

Using site-directed mutagenesis, you have created a mutant form of chymotrypsin that has alanine substituted for the usual serin

e at position 195. Which of the following effects would your expect to observe? a. No effect or slight increase in affinity for substrate coupled to a complete loss of enzyme activity. b. A decrease in the affinity for substrate coupled to a decrease in enzyme activity. c. An increase in the rate of peptide-bond cleavage due to an increase in the rate of acid-base catalysis. d. An increase in the rate of peptide-bond cleavage due to an increase in the rate of covalent catalysis. e. A complete loss of enzyme activity due to the inability to bind substrate
Biology
1 answer:
tankabanditka [31]3 years ago
4 0

Answer:

The correct answer is - option E.

Explanation:

Serine may not be affected by the mutation as it has no part of the hydrophobic binding site so it won't affect the affinity of the binding of the substrate. The peptide bond-breaking would not occur without the catalytic end or part so the enzyme would be the complete loss of enzyme activity due to site-directed mutagenesis that is created from the chymotrypsin.

Thus, the correct answer is - option E.

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Explanation:

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3 years ago
The following F2 results occur from a dihybrid cross (AaBb x AaBb): purple: A_B_ 9/16 white: aaB_ 3/16 white: A_bb 3/16 white: a
e-lub [12.9K]

Answer:

b. 3 (whilte) : 1(purple)

Explanation:

The first dihybrid crossing is between AaBb x AaBb

Now, if the double heterozygous traits self crossed, we have the following gametes shown below for the F₂ crossing.

AaBb = (AB, Ab, aB, ab)

                    AB                     Ab                     aB                     ab

AB                AABB                AABb                AaBB                AaBb

Ab                AABb                AAbb                AaBb                Aabb

aB                AaBB                 AaBb                aaBB                aaBb

ab                AaBb                 Aabb                 aaBb                aabb

We were being told that the results are;

purple: A_B_ 9/16

white: aaB_ 3/16

white: A_bb 3/16

white: aabb 1/16

From above, we can see that the same is true;

For Purple color; we Have (AABB and AaBb) since A is dominant to a and B is dominant to b.

∴ From the above punnet square; we have 9 purple colors which are:

(AABB, AABb, AaBB, AaBb, AABb, AaBb, AaBB, AaBb, AaBb) = 9/16

white: aaB_ ( since B is dominant to b, it can be either BB or Bb)

= (aaBB, aaBb, aaBb) = 3/16

white: A_bb ( since A is dominant to a, it can be either AA or Aa)

= (AAbb, Aabb, Aabb) = 3/16

white: aabb i.e homozygous recessive = (aabb) = 1/16

Furthermore, the question goes further by saying:

If a double heterozygote (AaBb) is crossed with a fully recessive organism (aabb), what phenotypic ratio is expected in the offspring?

If AaBb self crossed, we have:  (AB, Ab, aB, ab)

If aabb self crossed, we have: (ab, ab, ab, ab)

                    AB                     Ab                     aB                     ab

ab                AaBb                Aabb                aaBb                 aabb

ab                AaBb                Aabb                aaBb                 aabb

ab                AaBb                 Aabb               aaBb                 aabb

ab                AaBb                 Aabb                aaBb                aabb

AaBb ( purple)

= 4/16

= 1/4

= 0.25

= 25%

Aabb (white)

= 4/16

= 1/4

= 0.25

= 25%

aaBb (white)

= 4/16

= 1/4

= 0.25

= 25%

aabb (white)

= 4/16

= 1/4

= 0.25

= 25%

Now the proportion of white to purple =  (25%+25%+25%): 25%

=  75%:25%

= 3(white):1(purple)

We can therefore conclude that the expected phenotypic ratio in the cross between a double heterozygote (AaBb) with a fully recessive organism (aabb) yeids;

3 (whilte) : 1(purple)

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Answer:

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Explanation:

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Answer:

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