Given the information coding of DNA strand:
5'-TTT-TAC-GAA-GAG-TGA-3',
Write the corresponding DNA template and mRNA strand
DNA template: 3' ------ 5'
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mRNA Strand: 5'------3'
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in DNA 'A' base pair with 'T' and G base pair with C
while in mRNA A base pair with U
so
DNA strand 5' - 3'
5'-TTT-TAC-GAA-GAG-TGA-3',
DNA template: 3' ------ 5'
AAA-ATG-CTT-CTC-ACT
mRNA Strand: 5'------3'
UCA-CUC-UUC-GUA-AAA
Given the information coding of DNA strand: 5'-TTT-TAC-GAA-GAG-TGA-3', Write the corresponding DNA template and mRNA strand...
Type of DNA/RNA mutation: Type of protein mutation: 2. TTT 5' ATG 3" TAC GAG cтa GCT cGA CTT GAA GAA cтт AAA AAA ттт TAA ATT 3' 5 The corresponding mRNA would then be: And the protein that would be formed would be: Type of DNA/RNA mutation: Type of protein mutation:
You have the following mRNA sequence: 5’-GCAUGAUAUGCGAGCUAHCAUGACGU-3’ What is the DNA coding strand, template strand, and tRNA sequence. Where is the start and stop codons and provide the resultant polypeptide sequence
Here is a strand of mRNA: 5’-AAUCAUGUCCGAGGGCUACCCGUAG-3’ Write the sequence of the template strand of DNA that gave rise to this messenger RNA.
9. Transcribe the top strand of this DNA into RNA. Label the 3' and 5' ends. DNA: 3' TAC CCC GGG AAA TTT ACT 5' 5' ATG GGG CCC TTT AAA TGA 3' S'AnGGGG CCc nUM AAA UGA? 10. Now translate your mRNA into Protein. Label the C-term and N-term.
Transcribe the following DNA 1. 3’ – TAC CAC GTG GAC TGA GGA CTC CTC TTC AGA CGG CAA TGA – 5’ a. MRNA b. 5’ - AUG GUG CAC CUG ACU CCU GAG GAG AAG UCU GCC GUU ACU – 3’ 2. 5’ – AUG GUG CAC CUG ACU CCU GAG GAG AAG UCU GCC GUU ACU – 3’ a. PROTEIN b. 5’ - MET VAL HIS LEU THR PRO GLU GLU LYS SER ALA VAL THR – 3’...
Enter the corresponding section of mRNA produced from the following section of DNA template strand: 3' T T T G A A G C T C C A 5 Enter the nucleotide sequence using capitalized abbreviations. 5'_answer_3'
multiple choice question Consider the DNA coding (non-template) strand: 5-AAG TAC-3 What would be the amino acid sequence (using the three-letter abbreviations) for the resulting dipeptide? Becond GU الا لا 16 First Base BE LE ACC Third Base > COCO لا لا لا | GCA GCG GAG- GGG - Lys-Tyr Phệ-Met His-Glu Lys-Tyr 8 Phe-Met His-Glu Met-His + Change seat Send a message to the instructor < Join another session
Consider the most common Cystic Fibrosis variant: Wild-type CFTR DNA (Coding-strand): 3 ATC ATC TTT GGT GTT atc att ggt gtt... Cystic Fibrosis odi 1. Add the 5' and 3' designations to the DNA. 2. Write the template sequence for the DNA 3. Transcribe the wild-type and CF DNA into mRNA. Be sure to include the 5'/3' labels. 4. Translate the wild-type and CF mRNA into protein. Be sure to label these ends also. 5. In the above wild-type protein,...
PLEASE HELP WITH TABLE.thank you
1.Select the coding strand, and select the template strand from
the answers below. (Select more than 1 answer)
The coding strand is the first strand running from 5' to 3'.
The coding strand is the second strand running from 3' to
5'.
The template strand is the second strand running from 3' to
5'.
The template strand is the first strand running from 5' to
3'.
2.Given DNA sequence: 5’ TCCGATTGG 3’. Which of the...
Below is the template strand for a small gene. Draw the coding strand, the mRNA produced, and the polypeptide. 3’-TCCGTAACC-5’