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Leni [432]
2 years ago
9

Pls help me help me ​

Biology
2 answers:
Zolol [24]2 years ago
5 0

Answer:

Task card 25= B, sleet

Task Card 26= B, gasses

Task Card 27= A, Stratus

Task Card 28= B, warmer

eimsori [14]2 years ago
3 0

25: B

26: B

27: A

28:B

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Considering the same population of cats as in Part A, what is the expected frequency of each genotype (TLTL, TLTS, TSTS ) based
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Answer:

P = f(TLTL) = 0,16

H = f(TLTS) = 0,48

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Explanation:

Hello!

The allele proportion of any locus defines the genetic constitution of a population. Its sum is 1 and its values ​​can vary between 0 (absent allele) and 1 (fixed allele).

The calculation of allelic frequencies of a population is made taking into account that homozygotes have two identical alleles and heterozygotes have two different alleles.

In this case, let's say:

f(TL) = p

f(TS) = q

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Considering the genotypes TLTL, TLTS, TSTS, and the allele frequencies:

TL= 0,4

TS= 0,6

Genotypic frequency is the relative proportion of genotypes in a population for the locus in question, that is, the number of times the genotype appears in a population.

P = f(TLTL)

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Also P + H + Q = 1

And using the equation for Hardy-Weinberg equilibrium, the genotypic frequencies of equilibrium are given by the development of the binomial:

p^{2} = f(TLTL)

2pq = f(TSTL)

q^{2} = f(TSTS)

So, if the population is in balance:

P = p^{2}

H = 2pq

Q = q^{2}

Replacing the given values of allele frecuencies in each equiation you can calculate the expected frequency of each genotype for the next generation as:

f(TLTL) = P = p^{2} = 0,4^{2} = 0,16

f(TLTS) = H = 2pq = 2*0,4*0,6 = 0,48

f(TSTS) = Q = q^{2} = 0,6^{2} = 0,36

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