Answer:
Thymine in DNA occurs as the result of thymidylate synthase creating deoxythymidine monophosphate (dTMP), which then undergoes phosphorylation to deoxythymidine diphosphate (dTDP), then to Deoxythymidine triphosphate (dTTP), and incorporated into DNA by the DNA polymerase (DNA pol). Thymine in tRNA arises post-transcriptionally, by S-adenosylmethionine-dependent methylation of a uridine 5'-monophosphate (UMP) residue in RNA.
Explanation:
Thymidylate synthase is an enzyme involved in <em>de novo</em> DNA synthesis. This enzyme (thymidylate synthase) catalyzes the transfer of the one-carbon group from 5,10-methylene-tetrahydrofolate (5,10-CH2-THF) to deoxyuridine monophosphate (dUMP) and subsequent methylation to produce deoxythymidine monophosphate (dTMP), which is then phosphorylated to deoxythymidine triphosphate (dTTP) by kinases and incorporated into DNA. On the other hand, specific tRNA methylases catalyze the methylation of transference RNA (tRNA) by using S-adenosylmethionine as a methyl donor. Since tRNA methylation is a post-transcriptional modification, this chemical reaction is considered an epitranscriptomic modification on the RNA molecule.
Answer:
Since the child 4 years old WIC milk is better, it is highly nutritious, though cow milk may not has side effect on the child because at that age cholesterol cannot affect the child but help in brain development
Explanation:
The bond energy is defined as equal to the amount of energy given out when the bond is made (or absorbed when that the bond is broken).
So when a bond is made, and the system cools to its original energy, the amount of energy given out, -DeltaH, is more or less equal to the bond energy.
Noticed that minus sign. For a process that releases heat, DeltaH is negative.
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Hope this helps!
Answer: The probability of obtaining an individual who is CcmmLLPp 1/32.
Explanation: This can be achieved by crossing similar genes to obtain individual probability as shown in the attached image.
When you cross Cc in the first genotype with CC in the second genotype, the following probabilities will be obtained;
P (CC) = ¾, P(Cc) = ¼.
Similarly, crossing Mm with mm, we get;
P (Mm) = ½, P (mm) = ½
Crossing Ll with Ll, we get
P (LL) = ¼, P (Ll) = ½, P (ll) = ¼
Crossing PP with pp, we get
P (Pp) = 1
Therefore, the probability of individual with genotype CcmmLLPp will be;
P (Cc) x P (mm) x P (LL) x P (Pp)
= ¼ x ½ x ¼ x 1
= 1/32
