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UNO [17]
3 years ago
6

. Plants exchange gas with the atmosphere. Which statement ac​

Biology
1 answer:
joja [24]3 years ago
5 0

Answer:

Plants exchange two gases in the atmosphere i. e. oxygen and carbondioxide. Carbondioxide is entered into the plant body through small opening called stomata which are present on the surface of leaves. Carbondioxode is used as a raw material in the process of photosynthesis. Oxygen is produced in the end of photosynthesis process and releases in the atmosphere.

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Read the paragraph from the section "Pea Plants And The Punnett Square."
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A pea plant could have one copy of the height gene that coded for “tall” and one copy of the height gene that coded for “short”. The tall allele is “dominant”, and this means that it can mask the other allele.

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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

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brainly.com/question/419732?referrer=searchResults

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