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Read the scenario below and answer the questions that follow for 2 points each: Two Siamese and three Persian cats survive a shipwreck and are carried on driftwood to a previously uninhabited tropical island. All five cats have normal ears, but one carries the recessive allele f or folded ears, his genotype is Ff. The other four cats are not carriers, so their genotype is FF. Thus there are 10 total alleles within this population, 9 being F alleles and 1 being f. So, the frequencies are 9/10 or 0.90 for F the allele and 1/10 or 0.10 for the f allele. 1. Identify ONE of the five conditions required for Hardy-Weinberg equilibrium that this population of cats does not meet. 2. If you assume Hardy-Weinberg equilibrium for this population, what would be the frequencies of the two alleles after 10 generations? 3. Knowing that the frequency of the folded ears genotype ff is equal to to the frequency of the f allele squared (0.1 x 0.1), how many cats would have folded ears if after 10 generations there are 2,500 cats on the island?

Respuesta :

jebizimm26

Answer:

c

Explanation:

marianaegarciaperedo

According to H-W, a population is in equilibrium if it meets five conditions, and have the same allelic and genotypic frequencies through many generations. 1) Infinite population size / 2) p=0.9 and q=0.1 / 3) 25 cats

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Let us review some theoretical framework about Hardy-Weinberg equilibrium, while we answer the questions.

⇒ Hardy-Weinberg assumption for a population in equilibrium:

  • Random matings
  • No superposed generations
  • No mutations
  • No migration  
  • Infinite population size ⇒ No genetic drift
  • No natural selection

Q1. Identify ONE of the five conditions required for Hardy-Weinberg equilibrium that this population of cats does not meet.

Since this population is composed of two Siamese and three Persian cats, the condition that this population does not meet is infinite population size.

This population is under the effect of genetic drift for being only 5 individuals colonizing an island.                                    

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⇒ According to Hardy-Weinberg, p and q represent the allelic frequencies in a locus, referring to the allelic dominant or recessive forms.  

The genotypic frequencies after one generation are p² (H0m0zyg0us dominant), 2pq (Heter0zyg0us), q² (H0m0zyg0us recessive).  

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.
  • The genotypic frequencies sum also equals 1, this is p² + 2pq + q² = 1.

Being

  • p the dominant allelic frequency,
  • q the recessive allelic frequency,
  • p² the h0m0zyg0us dominant genotypic frequency
  • q² the h0m0zyg0us recessive genotypic frequency
  • 2pq the heter0zyg0us genotypic frequency

Q2. If you assume Hardy-Weinberg equilibrium for this population, what would be the frequencies of the two alleles after 10 generations?

Allelic frequencies, f(x), are

  • f(F) = p = 0.9
  • f(f) = q = 0.1

If this population is in H-W equilibrium, the allelic frequency will be the same through many generations.

So, after 10 generations, the allelic frequencies will still be p = 0.9 and q = 0.1.

Q3. Knowing that the frequency of the folded ears genotype ff is equal to to the frequency of the f allele squared (0.1 x 0.1), how many cats would have folded ears if after 10 generations there are 2,500 cats on the island?

We know that,

  • f(F) = p = 0.9
  • f(f) = q = 0.1
  • N = 2500 cats

To get the number of individuals with folded phenotype, we should get the phenotypic frequency, which equals the genotypic frequency because it is the recessive phenotype.

Genotypic and Phenotypic Frequency

  • F(ff) = q² = 0.1 = 0.01

Now that we have the phenotypic frequency, we just need to multiply it by the total number of individuals in the population, N.

N = 2500 individuals

  • Individuals with folded ears = 0.01 x 2500 = 25

So, after 10 generations and a population size of 2500 cats, there will be 25 folded-eared individuals.

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