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bulgar [2K]
3 years ago
14

If the PP of a test Protocol was 32 then the BP would have to be..

Biology
1 answer:
Harlamova29_29 [7]3 years ago
8 0

Answer:

The correct answer would be - 120/88 or 112/80

Explanation:

Pulse pressure or PP is the difference between the top or higher number (systolic) and the bottom number (diastolic) blood pressure. Pulse pressure can find out by the systolic minus the diastolic of Blood pressure.

For example, if an individual's resting BP is 120/80 millimeters of mercury (mm Hg), the normal PP is 40 which is considered a healthy pulse pressure. So, as per the normal BP 120/80, if the pp was 32 than the bp would be 120/88 or 112/80.

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Read 2 more answers
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 0
3 years ago
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