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2 years ago

Complete the comparison of the metabolic pathways leading to thymine in DNA and thymine as a modified base in tRNA.

Biology

1 answer:

asambeis [7]2 years ago

8 0

Answer:

The question is incomplete, the complete question is as follow:

Thymine in DNA occurs as the result of thymidylate synthase creating _______, which then undergoes _______ to _______, then to _______, and incorporated into D N A by _______. Thymine in tRNA arises ________, by an S-adenosylmethionine-dependent _______ of a ______ residue in RNA.

Answer:

Thymine in DNA occurs as the result of thymidylate synthase creating dTMP, which then undergoes phosphorylation to dTDP, then to dTTP, and incorporated into DNA by DNA polymerase. Thymine in tRNA arises posttranslationally, by an S-adenosylmethionine-dependent methylation of a UMP residue in RNA.

Explanation:

Thymidylate synthase is a key enzyme that catalyzes the synthesis of 2′-deoxythymidine-5′-monophosphate (dTMP) from 2′-deoxyuridine-5′-monophosphate (dUMP). This enzyme catalyzes the addition of a phosphate group (phosphorylation) of dTMP in order to form thymidine 5'-diphosphate (dTDP). Once synthesized, dTMP can be metabolized to deoxythymidine triphosphate (dTTP), which is a precursor for DNA biosynthesis. On the other hand, transfer ribonucleic acid (tRNA) is methylated after the molecule is synthesized. The transfer of the methyl group is catalyzed by an S-adenosylmethionine–dependent methyltransferase that catalyzes the methylation of uridine monophosphate (UMP) using SAM (S-adenosylmethionine) as the methyl donor.

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2 years ago

Calculate the final concentration of BSA in the problems below using the formula

Rudiy27

Answer:

A. $C_2=1.5\frac{mg}{mL}$

B. $C_2=0.075\frac{mg}{mL}$

C. $C_2=0.01\frac{mg}{mL}$

D. $C_2=0.001\frac{mg}{mL}$

Explanation:

Hello.

In this case, we must compute the final concentration in all the cases so we solve for it in the given equation:

$C_2=\frac{C_1V_1}{V_2}$

Thus, we proceed as follows:

A. Here, the final diluted solution includes the 300 μL of the 5 mg/ml-BSA and the 700 μL of TBS.

$C_2=\frac{300\mu L*5\frac{mg}{mL} }{(300+700)\mu L}\\\\C_2=1.5\frac{mg}{mL}$

B. Here, the final diluted solution includes the 50 μL of the 1.5 mg/ml-BSA, the 450 μL of water and the 500 μL of TBS.

$C_2=\frac{50\mu L*1.5\frac{mg}{mL} }{(50+450+500)\mu L}\\\\C_2=0.075\frac{mg}{mL}$

C. Here, the final diluted solution includes the 10 μL of the 1 mg/ml-BSA and the 990 μL of TBS.

$C_2=\frac{10\mu L*1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.01\frac{mg}{mL}$

D. Here, the final diluted solution includes the 10 μL of the 0.1 mg/ml-BSA and the 990 μL of TBS.

$C_2=\frac{10\mu L*0.1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.001\frac{mg}{mL}$

Best regards.

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2 years ago