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NeTakaya
3 years ago
11

Which gene may play a role in patterning the forelimb?

Biology
1 answer:
LUCKY_DIMON [66]3 years ago
3 0

Answer: Posterior HoxA and HoxD genes play important roles in patterning along the proximodistal axis of the forelimb (10–13). They also are required for activating and maintaining Shh expression in the developing forelimb (6, 7). mark brainliest

Explanation:

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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!

6 0
3 years ago
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