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3 years ago

transcribe the highlighted portion of DNA into mRNA transcription mRNA copies of DNA into RNA using complementary bases

Biology

2 answers:

Elina [12.6K]3 years ago

8 0

Ans. The correct answer is 'A-U-G-G-C-A.'

Transcription involves synthesis of mRNA molecule from a parent DNA molecule, with the help of RNA polymerase enzyme.

The RNA polymerase adds nucleotides in the growing RNA chain by reading template DNA strand, according to complementary base pairing, i.e., adenine binds with uracil, thymine binds with adenine, and cytosine binds with guanine, and guanine binds with cytosine.

Template DNA strand is : T-A-C-C-G-T

The mRNA strand will be: A-U-G-G-C-A.

ivann1987 [24]3 years ago

4 0

A - U - G - G - C - A

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Answer:

b. spores of two types to spores of one type

Explanation:

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Answer:

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3 years ago

10 points! When looking out of my back door, I see robins, blue jays, finches, crows, hawks and owls all living in the same area

forsale [732]

The possibility for all of the mentioned birds to live all in a same area is because they are all specialized and occupy a certain niche in the food chain, thus not standing in each others ways when it comes to competition for food.

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The robins and the blue jays are birds that are very opportunistic in their food choice, thus giving them greater flexibility and ability to survive, they mostly feed on worms, insects, fruits, vegetables, so they have a big menu.

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4 years ago

What do you mean by Goosebumps?? How it occurs

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Answer:

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3 years ago

Read 2 more answers

Since the llamas will not sell well with malformed feet, you decide to select the herd against this defect. You separate the aff

swat32

The question is incomplete as the frequency of alleles is not given, however, the frequency and population are given below :

Frequency of a = 0.506

total population = 500, Number of aa = 128

Answer:

The correct answer is - option C. 0.336.

Explanation:

Let, A = Normal allele, a = Defective allele

So, AA & Aa will develop normal phenotype & aa will develop defective phenotype.

Frequency of a = 0.506

So, Frequency of A = 1 - 0.506 = 0.494

So, frequency of AA = (0.494)2 = 0.244036

So, frequency of Aa = 2 x 0.494 x 0.506 = 0.499928

so, frequency of aa = (0.506)2 = 0.256036

total population = 500, Number of aa = 128

So, Number of AA = (0.494)2 x 500

= 122.018 which is almost 122 so considering it 122

So, Number of Aa = (2 x 0.494 x 0.506) x 500

= 249.964 which is almost 250 so considering it 250

It is given that, Relative fitness of AA (W11) & Aa (W12) is 1, and the relative fitness of aa (W22) is 0.

Now, mean fitness = (Frequency of AA x Fitness of AA) + (Frequency of Aa x Fitness of Aa) + (Frequency of aa x Fitness of aa)

= (0.244036 x 1) + (0.499928 x 1) + (0.256036 x 0) = 0.244036 + 0.499928 = 0.743964

So, after selection frequency of Aa

= (Frequency of Aa x Fitness of Aa) / mean fitness

= 0.499928 / 0.743964 = 0.67198 (Up to 5 decimal)

So, after the selection frequency of aa

= (Frequency of aa x Fitness of aa) / mean fitness

= 0 / 0.743964 = 0

So, frequency of a

= 1/2 of frequency of Aa + Frequency of aa

= 1/2 x 0.67198 + 0

= 0.33599 + 0

= 0.33599 which is almost 0.336

8 0

3 years ago