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dybincka [34]
3 years ago
10

Which individual are favored in natural selection?

Biology
1 answer:
Airida [17]3 years ago
4 0

Hello! :)

It would take forever listing all the animals that have adaptations that helps them survive in their environment.

Natural selection is the process by which individuals with characteristics that are advantageous for reproduction in a specific environment leave more offspring in the next generation, thereby increasing the proportion of their genes in the population gene pool over time.

Hope this helped and I hope I answered in time!

Good luck!

~ Destiny ^_^

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In a population of 10,000 individuals of Monstera deliciosa, four individuals have a variegated phenotype that is caused by the
Mariulka [41]

Under the assumption of this population being in Hardy-Weinberg equilibrium,<em> the </em><em>probability</em><em> for the</em><em> fixation</em><em> of the var </em><em>recessive allele </em><em>equals 0.02 = 2%</em>

--------------------------------

<u />

<u>Available data:</u>

  • N = 10,000 individuals ⇒ Large population
  • 4 individuals have a variegated phenotype
  • variegated phenotype is caused by the recessive var allele
  • No selection, no additional mutation ⇒ we can assume Hardy-Weinberg equilibrium

We need to know the probability for the fixation of the var allele.

We assume that the population is in Hardy-Weinberg equilibrium, so there should be no evolution.

Since the population is in H-W equilibrium, the allelic, genotypic and phenotypic frequencies will remain the same generation after generation.

Now, we will calculate the allelic and genotypic frequencies of variegated individuals in the population.

There are<u> 10,000 individuals</u>, and only <u>4 have variegated.</u>

So, the phenotypic frequency, F(Var) is 4/10,000 = 0.0004 =<u> 0.4%</u>

Since this is a recessive phenotype, this value equals the genotypic frequency, F(vv) = <u>0.4%</u>

Finally, we can get the allelic frequency by taking the square root of this value.

F(vv) = q² = 0.0004

f(v) = q = √0.0004 = 0.02 = <u>2%</u>

According to these calcs, the probability for the fixation of the var allele is <u>2%</u>, and the probability that all individuals express the variegated phenotype is <u>0.4%</u>.

------------------------------

You can learn more about Hardy-Weinberg equilibrium at

brainly.com/question/12724120?referrer=searchResults

brainly.com/question/9870841?referrer=searchResults

brainly.com/question/419732?referrer=searchResults

brainly.com/question/13603001?referrer=searchResults

brainly.com/question/5028378?referrer=searchResults

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