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Rus_ich [418]
3 years ago
12

Which of the following components is involved in the initiation of transcription?

Biology
1 answer:
sergey [27]3 years ago
8 0

Answer:

Promoter.

Explanation:

Transcription may be defined as the process of generation of the RNA molecule from the DNA template. The process of transcription includes initiation, elongation and termination.

Promoter may be defined as the region of DNA that are located in the upstream region. The initiation of transcription requires the promoter region that allows the binding of RNA polymerase at the initiation site.

Thus, the correct answer is option (3).

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13. What feature of this plant was least likely inherited from its parents?
rusak2 [61]

Answer: B

Explanation: That'd be based on how it's taken care of

6 0
2 years ago
These the flow of electrons (the current) and where some of the electrons' energy gets converted into heat.
Pavlova-9 [17]

Answer:

Insulators

Explanation:

The purpose of insulators are to convert the energy into thermal or heat energy

3 0
2 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
In what parts of the cycle do you think phosphorus spends the most time​
disa [49]

Answer:

what is a phosphorus

Explanation:

5 0
2 years ago
Write the procedure to observe human cheek cell​
scoundrel [369]

Answer:

To observe the cheek cell,

  • Take a tooth pick use its blunt side to scrap inside the mouth.
  • You will see some deposition on the blunt side of tooth pick, make a smear on the clean slide in the center using that tooth pick.
  • Add a drop of methylene blue solution and place a coverslip, make sure that bubbles are avoided i.e. coverslip should be placed in the inclined manner.  
  • Remove the excess solution and observe it under the microscope first under 4X and then under 10X.

Observation:  

  • The cells observed are squamous epithelial cells. The small blue dots seen inside will be the bacteria from our teeth and mouth.
4 0
3 years ago
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