Human ovaries play an important role in reproduction. Which part of a

An amateur horticulturalist breeds tulips in the hopes of obtaining a flower with unique and desirable traits. After several gen

Whitepunk [10]

Horticulturalist’s best option is clone the desired tulip so there will be no genetic variations.

Explanation:

Cloning is a process by which replica of organism is made without having any genetically variations. It is done carefully to avoid any variations in the office bring that may be related to physical or genetic structure.

The process of cloning should be done under the supervision of expert to avoid any error which could lead to the variation in genetic structure. It is very necessary for the amateur horticulturalist who desires to have dual apps without any genetic variation to perform cloning.

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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:

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