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Evgesh-ka [11]
3 years ago
9

What are the 4 ingredients in DNA??

Biology
1 answer:
maksim [4K]3 years ago
4 0
<span>adenine (A), cytosine (C), guanine (G) and thymine (T)
I think
</span>
You might be interested in
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
In one type of dog, black spots (B) are dominant over brown spots (b) and long tails (L) are dominant over short tails (l). Use
stiks02 [169]

Answer:

There is a 0% chance of getting the phenotype of brown spots and short tails.

Explanation:

Here is a Punnett square

             BL              Bl              BL               Bl

BL       BBLL          BBLl          BBLL          BBLL

Bl        BBLl            BBll           BBLl           BBll

bL       BbLL           BbLl          BbLL          BbLl

bl        BbLl            Bbll            BbLl          Bbll

8 0
3 years ago
Read 2 more answers
DNA is a double-stranded nucleic acid formed by millions of linked nucleotides that each contain the sugar deoxyribose, a phosph
larisa [96]

Answer:  the answer is a i just took this on a test today

Explanation:

7 0
3 years ago
Read 2 more answers
Select the correct statement about the special fetal blood vessels.
maks197457 [2]

The question is incomplete. The complete question is:

Question: Select the correct statement about the special fetal blood vessels.

A The umbilical vein becomes the ligamentum teres.

B The distal parts of the umbilical arteries form the superior vesical arteries.

C The hepatic portal vein forms from the umbilical artery.

D The fossa ovalis becomes the f oramen ovale.

Answer:

C) The umbilical vein becomes the ligamentum teres.

Explanation:

After birth, many vascular changes occur in the newborn child as the pulmonary, renal and digestive systems start functioning. As the umbilical cord is tied off and severed, the blood does not flow through the umbilical arteries. The umbilical vein in the child collapses but remains as the ligamentum teres. Ligamentum teres is a round ligament and serves to attach the umbilicus to the liver. It is present in the free border of the falciform ligament of the liver.

6 0
3 years ago
Adult male humans generally have deeper voices than do adult female humans, which is the direct result of higher levels of testo
Gnom [1K]

Answer:

a. sexual dimorphism was developing over time in these species.

Explanation:

Sexual dimorphism refers to the presence of observable distinct phenotypic differences between the males and females of the same species. In humans, adult males have a larger larynx which is responsible for their deep voice than females. Females have a relatively smaller larynx. Therefore, a trend toward decreasing larynx size in adult females and increasing larynx size in adult males in fossils represents the development of sexual dimorphism in these species.

7 0
3 years ago
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