What is another name for the dorsal surface of a starfish

Cual es la diferencia entre cigoto; embrión y feto

astra-53 [7]

Answer:

in spanish-Los términos embrión y feto se refieren al bebé en desarrollo dentro del útero de la madre. ... Un embrión se denomina feto a partir de la undécima semana de embarazo, que es la novena semana de desarrollo después de la fertilización del óvulo. Un cigoto es un organismo unicelular resultante de un óvulo fertilizado.

Explanation:

in english- The terms embryo and fetus both refer to the developing baby inside the mother's womb (uterus). ... An embryo is termed a fetus beginning in the 11th week of pregnancy, which is the 9th week of development after fertilization of the egg. A zygote is a single-celled organism resulting from a fertilized egg.

8 0

3 years ago

Each isotope has a special name derived from Latin (protium, deuterium, and tritium). What structural feature do these names ref

Eduardwww [97]

An isotope of any element is the same, with a variation in the neutrons of the nucleus.

The mass number change but the atomic number doesn't.

In this case, protium, deuterium, and tritium are all hydrogen isotopes.

Protium is 1H or Hydrogen-1 is without neutrons.

Deuterium is 2H or Hydrogen-2 has one neutron.

Tritium is 3H or Hydrogen-3 has two neutrons.

8 0

3 years ago

Chymotrypsin

Georgia [21]

Answer: d. all of the above

Explanation:

Chymotrypsin is produced by the Pancreas and has the function of digesting proteins.It consists of three polypeptide chains connected by two inter-chain disulfide bridges. Chymotrypsin exerts an important biological function which is the hydrolysis of proteins in the small intestine. The cleavage of peptide bonds is selective, occurring only on the carboxyl side of aromatic side chains of tyrosine, tryptophan and phenylalanine, and large hydrophobic radicals such as methionine. Chymotrypsin also hydrolyzes ester

3 0

3 years ago

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

Which of the following are factors that affect climate?

Ganezh [65]

Where are the options

4 0

2 years ago