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lilavasa [31]
3 years ago
14

Given the inefficiency of two of the pathways shown in Figure 2-4, what advantage could

Biology
1 answer:
grandymaker [24]3 years ago
5 0
I’m assuming this entire thing is cellular respiration. pathway a most likely represents alcoholic fermentation, B represent lactic acidosis fermentation, C probably represents aerobic cellular respiration. Alcoholic fermentation doesn’t occur in humans, it occurs in micro bacteria and yeast. Lactic acidosis fermentation occurs in humans when there is not oxygen. That’s why you breathe so heavily after sprinting. Cellular respiration just happens naturally. Idk if that’s what the question is asking so feel free to comment and if this helped, brain answer?
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In a hypothetical population of 2500 people, 2275 people have brown eyes and 225 people have blue eyes (the homozygous-recessive
aev [14]

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

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3 years ago
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