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erik [133]
3 years ago
6

During ketogenesis, the liver synthesizes ketone bodies that can be used as an energy source. Put the following statements regar

ding ketogenesis in the correct order, beginning with a stimulus for ketogenesis.
a. Low glucose levels stimulate the breakdown of fatty acids to acetyl-Cox
b. HMG-CoA Loses acetyl-CoA forming a four-carbon ketone body.
c. A condensation reaction with acetyl-CoA produces the six-carbon HMG-CoA.
d. Two acetyl-CoA condense to form the four-carbon acetoacetyl-CoA.
e. 3-Hydroxybutyrate and acetoacetate can cross tie blood-brain barrier to provide fuel for the brain,
Biology
1 answer:
mihalych1998 [28]3 years ago
8 0

The following statements regarding ketogenesis in the correct order, beginning with a stimulus for ketogenesis:

a) Low glucose levels stimulate the breakdown of fatty acids to acetyl-CoA.

d. Two acetyl-CoA condense to form the four-carbon acetoacetyl-CoA.

c. A condensation reaction with acetyl-CoA produces the six-carbon HMG-CoA.

b)  HMG-CoA Loses acetyl-CoA forming a four-carbon ketone body.

e). 3-Hydroxybutyrate and acetoacetate can cross tie blood-brain barrier to provide fuel for the brain.

Explanation:

Ketogenesis takes place in mitochondria of the hepatic cells when energy source as glucose is depleted.

When carbohydrate is in not present in the cell the AMPK regulatory protein gets activated which slow downs lipogenesis and activates malonyl CoA decarboxylase enzyme.

Ketogenesis take splace in following steps:

1) When no or low carbohydrate is there the the fatty acid breakdown takes place.

2)  Oxidation of fatty acid takes place to form acetyl CoA takes place.

3) Acetoacetylyl CoA formation takes place by enzyme thiolase action in which 2 acetyl CoA reacts.

4) A six carbon compound named 3-hydroxy-3-methylglutaryl-CoA takes place place. The enzyme working is HMG CoA synthase.

5)  A compound gets form which is Acetoacetate by releasing an acetyl CoA by the actIon of enzyme HMG CoA lyase.

6). Acetoacetate and 3-hydroxybutarate  are used to provide energy to the body and the brain.

The acetyl CoA released enters the TCA cycle.

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3 years ago
In a hypothetical population of 2500 people, 2275 people have brown eyes and 225 people have blue eyes (the homozygous-recessive
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Answer:

In the next generation of 4000 children, 1680 of them will be heterozygous for the eye colour.

Explanation:

There's a population of 2500, 2275 of with have brown eyes and 225 blue eyes. <u>Let's call the dominant allele associated with brown colour "B" and the recessive allele associated with blue colour "b"</u>. So the possible genotypes are BB, Bb and bb, being BB and Bb brown eyed individuals and bb blue eyed individuals.

If the population it's in Hardy-Weinberg equilibrium, it means genotypic and allelic frequencies don't change from one generation to the following.

From the information given, we can calculate both allelic and genotypic frequencies.

First, we know that the frequency of the genotype bb it's the amount of blue eyed individuals over the total population.

  • f(bb)=225/2500=0.09

Additionally we know the allelic frequencies can be related to the genotypic ones when the population it's in Hardy-Weinberg equilibrium. Particularly we can say:

  • f(bb)=[f(b)]^2 => f(b)=[f(bb)]^(1/2)= 0.3 <em>(square root of f(bb)).</em>

Also, we can calculate the frequency of the B allele, as the probability of all alleles of the gene sum 1. In other words:

f(b)+f(B)=1 => f(B)=1 - f(b) = 1 - 0.3 = 0.7

So far, we have calculated the allelic frequencies, f(b)=0.3 and f(B)=0.7.

Now we can calculate the genotypic frequencies, using the equations of the Hardy-Weinberg equilibrium.

  • f(bb)=[f(b)]^2 => f(bb)=0.3^2=0.09
  • f(Bb)=2*f(B)*f(b) => f(Bb)=2*0.7*03=0.42
  • f(BB)=[f(B)]^2 => f(BB)=0.7^2=0.49

Finally, knowing that there are 4000 children in the next generation, to know how many of them are expected to be heterozygous for the eye colour, we should multiply the number of children for the probability of being heterozygous for the eye colour (which is the genotypic frequency for the genotype Bb).

  • Nº of heterozygous individuals = f(Bb)*total population= 0.42*4000
  • => Nº of heterozygous individuals =1680

<em />

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<h3>What is a tropism?</h3>

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