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Whitepunk [10]
3 years ago
12

Binding of glucose to hexokinase causes a conformational change in the enzyme. This is an example of the __________ model of enz

yme catalysis.
Biology
1 answer:
irina [24]3 years ago
6 0

Answer:

Binding of glucose to hexokinase causes a conformational change in the enzyme. This is an example of the<u> induce-fit </u>model of enzyme catalysis.

Explanation:

The induce- fit model is generally the most accepted theory for enzyme catalysis. This theory states that the active site of an enzyme is not always a perfect fit for a substrate. The substrate induces changes in the active site so that it can fit into the active site. This theory is contrary to the theory of lock and key model, which stated that substrates exist as a perfect match for particular active sites of an enzyme.

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What’s meant by the term genetic variation?
Pepsi [2]

Answer:

Genetic variation is a term used to describe the variation in the DNA sequence in each of our genomes. ... Individuals of a species have similar characteristics but they are rarely identical, the difference between them is called variation

Explanation:

8 0
3 years ago
Read 2 more answers
Effector molecule binding changes the behavior of enzymes by altering the equilibrium between the tense (T) state and the relaxe
9966 [12]

Answer:

Explanation:

<h2>Homotropic effector-</h2>
  • \text{accounts for the sigmoidal nature of a velocity versus substrate concentration curve}
  • \text{An enzyme's substrate.}
<h2>Both -</h2>
  • \text{Works by altering T/R ratio.}

The phrase  \text{"accounts for the sigmoidal nature of a velocity versus }

is relevant and can be applied for homotropic effector molecules since the heterotropic effector molecules have the possibility and affinity to change the sigmoidal curve to a more potential hyperbolic curve contingent upon the allosteric effector to be positive or negative modulator.

The expression isn't relevant for both homotropic and heterotrophic effectors since the two of them can tie to the allosteric site of allosteric enzymatic compounds.

The phrase \text{"works by altering the T/R ratio"} is significant and can be applied for both homotropic and heterotropic effectors.

The expression \text{"an enzyme's substrate"} is significant and applied for homotropic effectors just as when substrate molecules tie to the allosteric site of enzyme then it is regarded as homotropic effectors. The heterotropic effectors are effectors apart from substrate molecules.

The phrase \text{"alters the} \  K_m \ \text{ of an enzyme"} is not applied and insignificant to none of the heterotropic or homotropic effector molecules since K_m is significant for the enzymes that obey the Michaelis-Menten equation, but allosteric enzymes do not obey the Michaelis-Menten equation. Homotropic and heterotropic effectors are viable and efficient for allosteric enzymatic chemicals that don't contain

4 0
3 years ago
You are a botanist looking at the plant color in a new breed of Snapdragon. The two parent plants are Blue and Green and when yo
irakobra [83]

Answer:

Incomplete dominance is the inheritance pattern where the dominant allele did not mask the recessive allele completely and form a mix of both alleles. Here the inheritance is the incomplete inheritance. The ratio of F2 generation is 1:2:1.

Given:

R1R1 = 42

R2R2 = 39

R1R2 = 86

Total R1 alleles = 2*42+86 = 170

Total R2 alleles = 2*39+86 = 164

Total alleles = 334

Frequency of allele R1 = 170/334 = 0.51

Frequency of allele R2 = 164 / 334 = 0.49

Expected number of each phenotype:

Total population = 167

Blue = R1R1 = 0.51 * 0.51 * 167 = 43.44

Green = R2R2 = 0.49 * 0.49 * 167 = 40.10

Cyan = 2*R1*R2 = 2*0.51*0.49*167 = 83.46

Phenotype     Observed(O)    Expected (E)    O-E      (O-E)2     (O-E)2/E

Blue                 42                   43.44               -1.44       2.0736     0.0477

cyan                86                   83.46                 2.54     6.4516     0.0773

green              39                    40.1                    -1.1       1.2100      0.0302

Total              167                     167                                               0.1552

Chi-square value = 0.155

Degrees of freedom = no. of phenotypes – 1

Df = 3-1 = 2

Critical value = 5.99

Chi-square value of 0.155 is less than the critical value of 5.99. So we accept the null hypothesis.

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3 years ago
Why do we need to destarch a plant
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In photosynthesis experiment, the effect of the light will be observe during the starch production- thus you need to have a plant without a starch to observe if the starch is produced during photosynthesis. Letting the leaves of the plant deprived from light for 48 hours will be forced to use its reserved starch making the storage of the starch decrease.
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In cell division, what must happen to the cell in interphase to allow it to continue to prophase?
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