Determine the complementary strand of DNA that forms on this template DNA fragment during replication: 5′GGTTTCTTCAAGAGA3′

As you should recall, DNA, when not being actively transcribed, has a double helical structure.  This portion of the DNA has had the two strands separated in preparation of transcribing for a needed protein.  The following is one of the two complimentary strands of DNA: 3' - AACCAGTGGTATGGTGCGATGATCGATTCGAGGCTAAAATACGGATTCGTACGTAGGCACT - 5'    Q:  Based on written convention, i.e. the 3'-5' orientation, is this the coding strand or the template   strand?  ______________________________ Q:  Assuming this strand extends from base #1 to #61 (going left to right), interpret the correctly transcribed mRNA and translated polypeptide for bases 24 - 47: mRNA:      ___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___- polypeptide chain:  ________--________--________--________--________--________--________--________

a. Replicate this sense strand to create a double-stranded DNA helix. Write your answers in CAPS LOCK with NO SPACES between the nucleotides - e.g. ATGCCGAG..... TGAGGATGAAACTCACACCGGGGCGCAGTTTGGCACTTAGATTCTTGTACACGACCTAGTATAACACAGTT complementary strand: b. Using this DNA double helix, express the gene – i.e. determine the resulting polypeptide sequence by using the correct reading frame. Write your answers using the three letter abbreviation for each amino acid. polypeptide sequence: does the sense strand DNA sequence have 5’ and 3’ UTR sequences? 5'UTR =  3'UTR =

Insulin is synthesized as preproinsulin, which has 81 amino acids. How many heterocyclic bases must be present in the informational DNA strand to code for preproinsulin (assuming no introns are present)?

If DNA segments changes from GCATAG to GCATGG, this is a: MRNA Codon/Amino Acid Chart First Base Second Base Third Base A G UUU1 Phenylalanine UCUT UAUT FTyrosine (Tyr) UAC UGU, U FCysteine (Cys) UGCJ UUCJ (Phe) UCC Serine (Ser) UCA U UGA - Stop A UUA1 FLeucine (Leu) UUG- UAA1 FStop UAGJ UCG- UGG - Tryptophan (Trp) G CUU CAU1 CCU1 CGUT Histidine (His) CAC- CUC CCC Proline (Pro) CCA CGC FLeucine (Leu) CUA FArginine (Arg) CGA CAA1 Glutamine A CAGJ (Glu) CGG- CUG- CCG- ACU AAUJ Asparagine AUUT AGUT FSerine (Ser) AGC- U AUC FIsoleucine (lle) ACC AACJ(Asn) Threonine A AUA- (Thr) AAA1 FLysine (Lys) AAG- АСА AGA, FArginine (Arg) AGG- A Start Methionine AUG- (Met) ACG- G GUU GCU, GAU1 Aspartic Acid GGU U GAÇJ(Asp) GUC Fvaline (Val) GUA GCC GGC G Alanine (Ala) GCA Glycine (Gly) A GAAJ Glutamic Acid GGA GAG- (Glu) GUG- GCG- GGG- G

This is part of the Escherichia coli DNA sequence that contains an inverted repeat. (Note: top strand is the coding strand). 5'-AACGCATGAGAAAGCCCCCCGGAAGATCACCTTCCGGGGGCTTTATATAATTAGC-3' 3'-TTGCGTACTCTTTCGGGGGGCCTTCTAGTGGAAGGCCCCCGAAATATATTAATCG-5' Draw the structure of hairpin loop that will be formed during the end of transcription.

What is the sequence of the newly synthesize DNA segment if the template strand has the following sequence:3'-ATGGCCTATGCGAT-5'

Match the enzyme on the left with its role in DNA replication DNA polymerase I helicase DNA ligase DNA polymerase III topoisomerase primase 72 W w# 3 E $ 4 R % 5 T A 6 MacBook Pro Y & 7 U * 8 replaces primers with DNA connects Okazaki fragments to form a continuous strand of DNA synthesizes short RNA fragments used to initiate DNA synthesis Uses the 3'OH of an RNA primer to synthesize the leading strand and Okazaki fragments keeps DNA from getting tangled up ahead of the replication fork "unwinds" the DNA double helix at the origin and replication forks 1 ( 9 X 0 0 P + 11 Next

Give the mRNA and amino acid sequence from this DNA strand: CCATTAACCTTACTGCTGGCTAAATTCGTTGCT

If the following nucleotide sequence, CTC/TGT/AAG/ACC/TTT experienced a mutation resulting in the deletion of the second cytosine in the first DNA triplet so the sequence is now CT_/TGT/AAG/ACC/TTT, what would be the amino acid sequence created from this mutated DNA strand? Table of mRNA codons UUA, UUG = leucine AGG, AGA = arginine %3D CAU, CAC = histidine GUU, GÜC, GUA = valine GAA. GAG=glutamic acid GCU, GUA, GUG = alanine GAU, GAC = asparagine GGU, GGC, GGA = glycine UCA, UCU =serine CGU, CGC, CGA = arginine