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3 years ago

_____ has provided the ability to regrow tissue for knee or hip joints.

1 answer:

6 0

The most appropriate answer would be Biotechnology. Biotechnology has provided the ability to regrow tissue for knee or hip joints. It uses biological systems that are believed to help the development of products that is reserved for specific use.

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How can injury to tissue prompt cells in the surrounding area to undergo cell division?

Alona [7]

Cell division is the process essential for growth and repair in multi-cellular organisms.

Injury to the tissue can activate surrounding cells to undergo cell division with the help of mitogens. Mitogens are the chemical substances which help in escalating mitotic divisions in the cells.

They act by activating the signal-transduction pathway which involve mitogen-activated protein kinase (MAPK).

Other possible manner could be the loss of density which could lead to the inactivation of contact inhibition. As there is a loss of cells during the injury, it makes an empty space which could also trigger or activate surrounding cell to start cell division.

In addition, at injury site the flow of blood also increases to transport various nutrients, clotting factors, growth factors etc. These increased nutrition and growth factors also speed up the process of mitosis.

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3 years ago

What is the role of rna polymerase in transcription?

mojhsa [17]

To unwrap the DNA apart

7 0

3 years ago

Individuals with eating disorders share many of the same thought patterns as people with which condition?'

Klio2033 [76]

DEPRESSION. Eating disorder and depression goes hand in hand which also gives almost the same symptoms such as sadness, agitated, sleep problems, loss of interest. Treatment for this condition must be tailored specifically to the person and must be evaluated properly.

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).

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3 years ago

Some examples of non-polar molecules include fats, oils, and waxed. How do these substances interact with water?

77julia77 [94]

Since non-polar molecules are equally distributed and positively charged atoms in contrast with polar molecules. Thus, like fats oils and waxes, these materials are composed mainly of lipids or in the group of fats and are conspicuously insoluble with water. How do these substances interact with water? They are hydrophobic, thus, they don't "mix" with water. Unlike water and other polar molecules that are unequally and negatively distributed and charged making them able to bond with other molecules like oxygen and other polar molecules and substances.

6 0

3 years ago