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2 years ago

Please helpppppp will give brainliest

Biology

1 answer:

Furkat [3]2 years ago

3 0

Answer:

Embryotic stem cell research benefits our society with the possibility of it helping various diseases and it only requires a small number of cells. The negatives of stem cell research are the cost and ethical controversary. To start, stem cell research could help with diseases like Alzheimer's and certain cancers like Parkinson's. Also since cells replicate incredibly fast, procedures wouldn't need many cells depending on the case. Stem cell research can vary in costs from about $5000 to $20,000 which for many people is difficult because stem cells are very temporary in many cases. Many groups of people don't agree with stem cell research since human embryos are destroyed in the process of harvesting embryonic cells and it is "potential life".

Thats all i could do i hope this helps

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Many bedbugs are highly resistant to insecticides, but the scientist’s bedbugs aren’t resistant. Explain why this happens based

Nutka1998 [239]

Wild Bedbugs become insecticide resistant because of the mutations and natural selections.

<h3><u>Explanation</u>:</h3>

As the huge amount of pesticides and insecticides are sprayed in the rooms for cleaning, the pests and insects like bedbugs dies in huge portions because of the toxin. But some of the bedbugs remain alive as they have mutations that help them to detoxify the toxins given, or bypass the metabolic processes so that the toxins don't hamper them much.

Now as the population becomes very small(bottle neck effect), the nature selects these organisms over the other to propagate more sufficiently and enormously. As the nutrients and supplies are also available, so the bedbugs don't suffer any lack of nutrition which can be a determining factor of their population.

Thus the wild bedbugs become resistant to insecticides while the experimental one remain succeptible to insecticides.

4 0

3 years ago

PLS HELP!!!

choli [55]

D. Oxygen and metal

They usually contain some form of metallic cation, given that carbonates are the most distributed minerals in the Earth's crust.

Hope this helped!

5 0

3 years ago

Read 2 more answers

A scientist isolates mRNA from the cytoplasm of some mouse cells. She separately isolates the DNA from mouse cells and heats the

tatyana61 [14]

Answer:

Answer is explained below.

Explanation:

A. The observed single stranded regions are found in the mRNA.

B. The loops represent introns (Non-coding portions of the mRNAin primary transcript ). The intron sequences are removed to form a mature mRNA by splicing.

C. If the scientist use RNA and DNA from bacteria, loops cannot be seen. Because introns are produced in only eukaryotes but not in prokaryotes.

5 0

3 years ago

Read 2 more answers

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

Suppose a segment of mitochondrial DNA (mtDNA) is compared between two similar modern-day species. It is known that this segment

QveST [7]

Answer:

24 million years

Explanation:

A: GCACTAAGCATCGATTT

B: GCACCAGGCACTGGTTC

There are 6 base pair changes between species A and species B. Since we know the rate of change is 1 base pair every 4 million years, we know 6x 4 million is likely how long ago the species diverged. 6x 4 million = 24 million years

7 0

3 years ago