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2 years ago

Please help quickly

Biology

2 answers:

blsea [12.9K]2 years ago

7 0

Answer:

use the camera lol....

USPshnik [31]2 years ago

7 0

Light availability determining the growth rate for plants

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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

zaharov [31]

Answer:

P = f(TLTL) = 0,16

H = f(TLTS) = 0,48

Q = f(TSTS) = 0,36

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

p + q = 1

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)

H = f(TLTS)

Q = f(TSTS)

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:

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

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

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

I hope you have a SUPER day!

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