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Elenna [48]
2 years ago
9

Martin is studying sucrase activity. He added sucrose solution + sucrase solution + buffer of pH 5.6 in two tubes. He incubated

them at 350C and 1000C respectively. When he performed a confirmatory test, only the first test tube showed enzyme activity. How would you justify the inactivity of the enzyme in the second test tube?
The enzymes in the second test tube must have denatured at pH 5.6, since most enzymes work at neutral pH.



The enzymes in the second test tube must have denatured at 1000C, since most enzymes degrade at such high temperatures.



Enzymes must have remained dormant at 1000C, since most of the enzymes remain inactive at such higher temperatures.
Biology
2 answers:
kodGreya [7K]2 years ago
4 0

Answer:

The correct answer is "The enzymes in the second test tube must have denatured at 1000C, since most enzymes degrade at such high temperatures".

Explanation:

Most enzymes are purified or produced from amino acid sequences found in nature. Since most organisms live in mild temperature conditions, enzymes are active under mild temperature conditions. This is the case of sucrase, which has activity at 35 celsius but at 100 celsius would not have activity as a result of temperature denaturation. This explains why the second test tube shows no enzyme activity, sucrase would must have denatured at 100 celsius.

lukranit [14]2 years ago
3 0
<span>B, enzymes will denature at that temperature. Sucrase has an optimal temperature of about 37 degrees celsius so 1000 would definitely be too high for them to work.</span>
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Answer:

D. 1 BB : 2 Bb : 1 bb

Explanation:

This question involves a single gene coding for fur color in guinea pig. Black fur (B) is dominant over white fur (b). This means that, as stated in the question, if a black fur parent (BB) is crossed with a white fur parent (bb), a 100% heterozygous offspring (Bb) with black fur will result.

If two heterozygous guinea pigs are crossed i.e. Bb × Bb, the following gametes will be produced by each heterozygous parent:

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Using these gametes in a punnet square (see attached image), offsprings with the following genotypic ratio will be produced:

1 BB : 2 Bb : 1 bb

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Answer:

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The fusion of these abnormal gametes produced by nondisijuction with a normal gamete (X or Y chromosome). This leads to a condition called ANEUPLOIDY, which is a chromosomal disorder. In the case od Calix, XXY is an ANEUPLOIDY state that could have resulted because there was a fusion between a XX gamete and Y gamete or XY and X. The XY or XX gametes are produced as a result of nondisijuction.

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