Transcribe the following Strand of DNA:

CCGATAGGT

Explanation:

got this for my hw.

So the central dogma of molecular biology describes the journey from DNA to protein product:

DNA --transcription--> mRNA --translation--> Protein

Assuming the DNA sequence provided is the template strand (rather than the complimentary coding strand), we start by transcribing the sequence into mRNA starting on the 3' end of the DNA towards the 5' end (which would build the mRNA 5' to 3'). This process involves the enzyme "RNA polymerase," which can only add nucleotides to the 3' end of the mRNA, just like how DNA polymerase can only synthesize DNA in the 5' to 3' direction. The RNA polymerase will bind to the template DNA strand and synthesize the complimentary mRNA, substituting uracil for thymine (since RNA does not contain thymine like DNA).

In terms of transcribing the sequence given to you, we'll have to work backwards + flip it around to get the 5' to 3' mRNA since the DNA is given 5' to 3' rather than 3' to 5'. Due to the length and the fact that we'll have to use triplets in translation anyways, it can help to break the sequence into triplet codons now.

5’-AAG | TTA | ATG | AGA | AAT | CGA | CAT | GGG | GCG | CCG | AAA | GTA | TAA | CCG | TCT | TAG | AAT | AGC-3’

We can then cross out each codon as we transcribe it and flip the sequence to be 5'-3' mRNA:

mRNA: 5'- GCU | AUU | CUA | AGA | CGG | UUA | UAC | UUU | CGG | CGC | CCC | AUG | UCG | AUU | UCU | CAU | UAA | CUU -3'

Normally, mRNA sequences start with "AUG" which is the start codon (and codes for Methionine), but I'll assume this is just for practice translating + transcribing in general. There's also a stop codon before the end but I'll assume the same again.

Translation involves three main steps - initiation, elongation, and termination. Initiation involves the translation ribosome assembling around the mRNA starting at the 5' end start codon, and tRNA carrying an amino acid binding to the complimentary section of the mRNA. As each tRNA attaches and the ribosome moves along the mRNA, the amino acids on each tRNA are bonded into a longer and longer peptide chain and the now amino acid-less tRNA are ejected (elongation). Termination occurs when a stop codon is reached, the ribosome will end elongation and help fold the protein into its final structure.

To translate the mRNA sequence here we'll need an amino acid/mRNA codon chart. I don't believe I can attach an image here, but looking up those exact words should yield the right results in images.

5'- GCU | AUU | CUA | AGA | CGG | UUA | UAC | UUU | CGG | CGC | CCC | AUG | UCG | AUU | UCU | CAU | UAA | CUU -3'

Ala - Ile - Leu - Arg - Arg - Leu - Tyr - Phe - Arg - Arg - Pro - Met - Ser - Ile - Ser - His - STOP - Leu

Amino acids are often abbreviated into three letters (Ala = alanine, Met = methionine, etc), and sometimes are abbreviated as single letters, though I've only seen that for sequencing databases.

In terms of locations for each of these processes, transcription occurs in the nucleus for eukaryotes and translation in the ribosomes/cytoplasm.

Explanation:

n

WIP inventory       68,000 debit

Factory Overhead   8000 debit

     Raw Materials Inventory   76,000 credit

WIP inventory         55.000 debit

Factory Overhead  12,400 debit

  Factory Payroll payable       77,400 credit

WIP inventory       31,250 debit

   Factory overhead      31,250 credit

Finished Goods Inventory 73,750 debit

           WIP inventory              73,750 credit

Explanation:

Direct Materials used:

10,000 + 20,000 + 24,000 + 14,000 = 68,000

Direct Labor used:

8,000 + 17,000 + 18,000 + 12,000 = 55,000

Overhead Applied:

6,000 + 12,750 + 13,500 + 9,000 = 31,250

Overhead rate:

6,000 /  8,000 =  0.75

12,750 / 17,000 =  0.75

Finished goods:

24,000 + 49,750 = 73,750