Females can get an X-linked recessive disorder, but this is very rare. An abnormal gene on the X chromosome from each parent would be required, since a female has two X chromosomes. ... In each pregnancy, if the mother is a carrier and the father has the disease, the expected outcomes are: 25% chance of a healthy boy.

Explanation:

3 0

3 years ago

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, the probability for the fixation of the var recessive allele equals 0.02 = 2%

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Available data:

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 10,000 individuals, and only 4 have variegated.

So, the phenotypic frequency, F(Var) is 4/10,000 = 0.0004 = 0.4%

Since this is a recessive phenotype, this value equals the genotypic frequency, F(vv) = 0.4%

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 = 2%

According to these calcs, the probability for the fixation of the var allele is 2%, and the probability that all individuals express the variegated phenotype is 0.4%.

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3 0

3 years ago

PLEASE HELP!! ASAP!

hammer [34]

A.) How tall the slope is

4 0

4 years ago