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Temka [501]
3 years ago
12

During embryonic development, if the anus develops before the mouth, the organism is a _____.

Biology
1 answer:
aliya0001 [1]3 years ago
8 0
<span>The question says, 'during embryonic development, if the anus develop before the mouth, the organism is a ............ The correct option is B. In deuterostome during embryonic development anus develops first from or near embryonic blastopore, mouth develops later at some distance from the blastopore or from the opposite end.</span>
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A particular recessive genetic disorder is fatal before birth, so there are no homozygous recessive individuals. In a particular
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Answer:

  • the allelic frequency for p is 0.967
  • the allelic frequency for q is 0.033

Explanation:

According to Hardy-Weinberg, the allelic frequencies in a locus are represented as p and q, referring to the alleles. The genotypic frequencies after one generation are p² (Homozygous for allele p), 2pq (Heterozygous), q² (Homozygous for the allele q). Populations in H-W equilibrium will get the same allelic frequencies generation after generation. The sum of these allelic frequencies equals 1, this is p + q = 1.

In the exposed example,

  • A recessive genetic disorder is fatal before birth, so there are no homozygous recessive individuals
  • In a particular population, one in 15 individuals is a carrier for this disorder.

What are the allele frequencies of the dominant (p) and recessive (q) alleles in this population?

If 1 of 15 individuals are carriers for this disorder, this means that 1/15 are heterozygous, 2pq. So, 2pq = 1/15 = 0.066

Now we must calculate the allelic frequencies.

We know that 1 in 15 individuals are heterozygous, and we also know that there are no recessive homozygous, q², because they can not survive, so of the 15 individuals only one is heterozygous and the rest 14 individuals must be dominant homozygous, p².

The dominant homozygous genotypic frequency is

p²= 14/15 = 0.933

And by clearing the next equation we can get the allelic frequency for p

p²= 0.933

p = √0.933

p = 0.967

So now we know that the allelic frequency for p is 0.967  

This means that the allelic frequency for q or p is 0.033, which we deduce by clearing the equation p + q = 1

                          0.967 + q = 1

                         q = 1 - 0.967

                          q = 0.033

  • the allelic frequency for p is 0.967
  • the allelic frequency for q is 0.033
5 0
3 years ago
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