Plz help!<br> I well mark brainliest!!

Cystic fibrosis (CF) is one of most common recessive disorders among Caucasians it affects 1 in 1,700 newborns. What is the expe

Phantasy [73]

Answer: The expected frequency of carriers is P(Aa)=0.046.

The proportion of childs with CF is P(aa)=0.024.

25% of having a child with CF (aa).

Explanation:

Hardy-Weinberg's principle states that in a large enough population, in which mating occurs randomly and which is not subject to mutation, selection or migration, gene and genotype frequencies remain constant from one generation to the next one, once a state of equilibrium has been reached which in autosomal loci is reached after one generation. So, a population is said to be in balance when the alleles in polymorphic systems maintain their frequency in the population over generations.

Given the gene allele frequencies in the gene pool of a population, it is possible to calculate the expected frequencies of the progeny's genotypes and phenotypes. <u>If P = percentage of the allele A (dominant) and q = percentage of the allele a (recessive)</u>, the checkerboard method can be used to produce all possible random combinations of these gametes.

Note that p + q = 1, that is, the percentages of gametes A and a must equal 100% to include all gametes in the gene pool.

The genotypic frequencies added together should also equal 1 or 100%, and all the equations can be summarized as follows:

$p+q=1\\(p+q)^{2} = p^{2} +2pq+q^{2} = 1\\P(AA)=p^{2} \\P(aa)=q^{2} \\P(Aa)=2pq1$

So, there are 1700 individuals and only one is affected. Since it is a recessive disorder, the genotype of that individual must be aa. So the genotypic frequency of aa is 1/1700=0.000588.

Then, $P(aa)=q^{2}=0.000588$ . And with that we can calculate the value of q,

$P(a)=q=\sqrt{0.000588}=0.024$

And since we know that p+q=1, we can find out the value of p.

$p+0.024=1\\1-0.024=p\\p=0.976$

Next, we find out the genotypic frequency of the genotype AA:

$P(A)=p=0.976\\P(AA)=p^{2} = 0.976^{2}=0.95$

Now, we can find out the genotypic frequency of the genotype Aa:

$P(Aa)=2pq=2 x 0.976 x 0.024 = 0.046$

Notice than:

$p^{2} + 2pq + q^{2} = 1\\x^{2} 0.976^{2} + 2 x 0.976 x 0.024 + 0.024^{2} = 1$

Then, the expected frequency of carriers is P(Aa)=0.046

The proportion of childs with CF is P(aa)=0.024

If two parents are carriers, then their genotypes are Aa.

Gametes produced by them can only have one allele of the gene. So they can either produce A gametes, or a gametes.

In the punnett square, we can see that there genotypic ratio is 2:1:1 and the phenotypic ratio is 3:1. So, there is a probability of 25% of having an unaffected child, with both normal alleles (AA); 50% of having a carrier child (Aa) and 25% (0.25) of having a child with CF (aa).

5 0

3 years ago

Year-old male with an extensive cardiac history presents with 2 hours of crushing chest pain and shortness of breath. he is pale

dezoksy [38]

Administer fluids to podsibly flush out viagra?? im nit 100% sure but it sounds like he could have taken too much

4 0

3 years ago

In what way do two alleles for the same trait differ?

kvasek [131]

I think it’s C explanation : i learned this a while ago

8 0

3 years ago

Which molecule is a product of photosynthesis?

zavuch27 [327]

Answer:water and sugar(also known as glucose)

Explanation:During the process of photosynthesis plants break apart the reactants of carbon dioxide and water and recombine them to produce oxygen(o2) and a form of sugar called glucose (C6H12O6)

5 0

2 years ago

Read 2 more answers

MUST BE at least 350 WORDS 50 POINTS

Alona [7]

Answer:

Sickle cell disease (SCD) affects millions of people around the globe and is the 4th leading cause of deaths in children in many developing countries. It causes a number of health problems, such as attacks of pain, anaemia, swelling in the hands and feet, bacterial infections and stroke. Sickle-cell contributes to a low life expectancy in the developed world of 40 to 60 years.

The disease results in abnormal haemoglobin - the oxygen-carrying protein found in red blood cells – giving the blood cell a rigid, sticky, sickle-like shape that hinders its oxygen-binding properties. These irregularly shaped cells can get stuck in small blood vessels, which can slow or block blood flow and oxygen to parts of the body. A blood and bone marrow transplant is currently the only cure for sickle cell disease, but only a small number of patients are eligible. For the rest, there's no cure but effective treatments can relieve pain, help prevent problems associated with the disease and prolong life.

70 years ago, researchers found a genetic connection to the anatomical abnormalities seen in blood cells. A mutation seemed to be causing the moon-shaped blood cells. The most severe form of the disease occurs when two copies of the mutation are inherited. However, patients with one sickle cell gene, referred to as sickle cell trait, usually do not have any of the signs of the disease and live a normal life, but they can pass the trait on to their children.

As with all inherited genetic diseases, you’d expect natural selection to weed out a gene that has such unpleasant consequences but with sickle cell disease, that doesn’t seem to be the case. Indeed, as of 2015, about 4.4 million people have sickle cell disease, while an additional 43 million have sickle cell trait. So what makes the disease stay in the human population?

Researchers found the answer by looking at where the disease was most prevalent. As it turns out, 80% of sickle cell disease cases occur in Sub-Saharan Africa or amongst populations having their ancestors in this region, as well as in other parts of the world where malaria is or was common. There was a long standing theory that the sickle cell trait – having only one sickle cell gene – didn’t cause discomfort and provided a bonus trait of preventing patients from contracting severe forms of malaria. Later confirmed - associating sickle cell to a 29% reduction in malaria incidence - this working theory would explain why the mutation stuck around in evolution. In 2011, researchers used mice to confirm the assumption.

Miguel Soares and Ana Ferreira of the Gulbenkian Institute of Science in Oeiras, Portugal, and colleagues found that haem – a component of haemoglobin – is present in a free form in the blood of mice with sickle cell trait, but largely absent from normal mice. By injecting haem into the blood of normal mice before infecting them with malaria, researchers found it could help guard against malaria. The mice did not develop the disease. Their results also showed that the gene does not protect against infection by the malaria parasite, but prevents the disease taking hold after the animal has been infected.

Explanation:

Sorry if I did or got anything wrong:(

I actually tried on this tho:)

3 0

3 years ago

Plz help! I well mark brainliest!!