The range or area occupied by a population is called

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

3 paragraph summary (I'll give brainlist if you do it correctly)

BARSIC [14]

Answer:

alright bet I got this

Explanation:

Have you ever wondered what can happen to those plastic bottles you toss away? How do they decompose? What happens to them? Its simple really, they get eaten! No, not by people of course they are eaten by a tiny critter called a wax worm. The wax worms earn their name because they are often found living in beehives where they eat the beeswax.

Wax worms much like mealworms are used to break down things that would normally be endangering to our environment. These seemingly useless little animals do a whole lot for our environment especially with the amount of waste we produce! They really are a tiny secret hero to todays pollution problem.

After quite a few studies scientists are still unsure on how was worms can breakdown plastic with their tiny bodies but they appreciate the work these little recyclers do. They have a lot of theories on how these worms do this one of the theories is that bacteria can break it down, another theory is that enzymes in the worms intestines break down plastics. Though scientists believe it will take a while to figure this out they enjoy studying and learning more about this tiny helper.

6 0

2 years ago

What are the two criteria for designation as a biodiversity hotspot?.

Ivan

Answer:

Biodiversity hotspots are regions under surveillance of humans where threatened and endemic species of flora and fauna are grown. These regions have been identified as some of the world's most important ecosystems.

The two criteria for designation as a biodiversity hotspot are:

1. It must possess at least 1500 vascular plant species as endemics : It means the region must have a high percentage of plant species which are found no where else in the world.

2. It must have 30% or less native vegetation : It means that the plant species population is endangered and threatened to extinction.

3 0

3 years ago

An example of sustainable resource use is the use of predators and parasites to A. Harm natural resourcesB. Pollinate plantsC. C

Ulleksa [173]

control pest insects

3 0

3 years ago

What is the main force by which most of the water within xylem vessels moves toward the top of a tree??

xeze [42]

Evaporation of water thrrough the Stoma

6 0

3 years ago