Write the corresponding section of a mRNA produced from each of the following DNA sections: a. DNA: 3'-A-T-G-A-C-G-C-T-A-5' b. DNA: 3'-G-T-A-C-C-A-T-A-C-G-A-G-G-C-5'
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- The following segment of DNA codes for a protein. The uppercase letters represent exons. The lowercase letters represent introns. The lower strand is the template strand. Indicate the 3’ and 5’ ends of both strands. G C T A T A A T G G C A a a a t t g G G T C A G G C A a a t c g a C A T A G C T G A C G G g g a t g a G G T T A A C G A T A T T A C C G T t t t a a c C C A G T C C G T t t a g c t G T A T C G A C T G C C c c t a c t C C A A T T 2.Write the pre-mRNA molecule. Indicate the 3’ and 5’ ends. 3. Write the mRNA molecule. Indicate the 3’ and 5’ ends 4. Write the tRNA anticodons corresponding to the codons in the mRNA. 5. Write the sequence of amino acids in the resulting polypeptide.The DNA sequence below is transcribed from left to right (the partner/coding strand is shown). Using this sequence, write the sequence of the polypeptide that results from this gene. Be sure to appropriately label the ends of the molecule. 5'-ATGCACGGCGACTAG-3' Second letter A UAU Tyr UAC First letter U A G U UUU1 UUC UUA LOU Leu CUU CUC CUA CUG Phe GUU GUC GUA GUG Leu AUU AUC lle AUA AUG Met Val C UCU UCC UCA UCG CCU CCC CCA CCG ACU ACC ACA ACG GCU GCC GCA GCG Ser Pro Thr Ala CAU His CAC CAA CAG Gin AAU Asn AAC AAA 1 Lys AAG LYS G {}a UAA Stop UGA Stop A UAG Stop UGG Trp GAU 1 GAC Asp GAA GIU Glu GAGJ UGU UGC CGU CGC CGA CGG AGU AGC AGA AGG Cys GGU GGC GGA GGG Arg Ser Arg DOA DOA DOA DUTO Third letter GlyIf the sequence of amino acids encoded by a strand of DNA is serine-alanine-lysine-leucine, what is the order of bases in the sense strand of DNA? Use the codon chart below to help you: second letter A G UAU Tyr UGU UUU UCU Phe Сys UUC UCC UAC UGC Ser UAA stop |UGA stop | A UAG stop UGG Trp UUA UCA UUG Leu G UCG CUU CCU CAU CGU His CUC ССС САС CGC Leu Pro Arg CUA ССА САА CGA Gln CUG CCG CAG CGG AUU ACU AAU AGU Asn Ser AUC le A AUA AAC AAA AGC AGA Arg АСС Thr ACA AUG Met | ACG AAG Lys AGG GUU GCU GAU GGU Asp GUC Val GUA GAC S GAA Glu GCC GGC Gly GGA Ala GCA GUG J GCG GAG GGG) O 3' AGACGTTTCAAT 5' O 3' UGUGCAAAGUUA 5' О 5 TGTGCTTТCТТА 3' first letter UCAG UCAG PCAG third letter
- A segment of a polypeptide chain is Arg-Gly-Ser-Phe-Val-Asp-Arg. It is encoded by the following segment of DNA: G G C T A G C T G C T T C C T T G G G G A C C G A T C G A C G A A G G A A C C C C T Template strand with its polarity: 3’ C C G A T C G A C G A A G G A A C C C C T 5’ - Coding strand with its polarity: 3’ G G C T A G C T G C T T C C T T G G G G A 5’ Please write out the mRNA sequence generated by the template strand to produce that polypeptide chain.Compare the two DNA sequences shown below and consider the single nucleotide mutation made in the lower DNA sequence (shown in bold font). This is an example of a mutation. DNA: ATG CGC ТСС САТ стт ААС АAА GAG GTT GG C TAT TT Protein: Met-Arg-Ser-His-Leu-Asn-Lys-Glu-Ala-Gly-Tyr-Phe DNA: АTG CGC ТСС САТ стТ ААС АAG GAG GTT GGC ТАТ ТТT Protein: Met-Arg-Ser-His-Leu-Asn-Lys-Glu-Ala-Gly-Tyr-Phe missense nonsense frame-shift silent antisenseWrite the base sequence that would be sticky with the sequence T-A-T-G-A-C-T.
- Below is the 5’–3’ strand of a double-stranded DNA molecule with the following nucleotide sequences:5’ C C T A T G C A G T G G C C A T A T T C C A A A G C A T A G C 3’1. If the above DNA strand is the coding (sense) strand and the DNA molecule is transcribed, what is the correct nucleotide sequence and direction of the RNA formed after transcription?The template strand of a segment of double-helical DNA contains the sequence – 5’-CTT-AAC-ACC-CCT-GAC-TTC-GCG-CCG-CAT-3’ a. What is the base sequence of the complementary strand of DNA? Indicate the 5’ and the 3’ ends. b. What is the base sequence of the mRNA that can be transcribed from this template DNA strand? Indicate the 5’ and the 3’ ends. c. What amino acid sequence can be coded by the mRNA in (b) starting from the 5’ end (or the N terminal amino acid)?Translate the following mRNA nucleotide sequence into an amino acid sequence, starting at the first base: 5’ - UGUCAUGCUCGUCUUGAAUCUUGUGAUGCUCGUUGGAUUAAUUGU - 3’
- Compare the two DNA sequences shown below and consider the single nucleotide mutation made In the lower DNA sequence (shown in bold font). This is an example of a mutation. DNA: АTG CGC TСС САТ СТТ ААС AAА GAG GTT GGC ТАТ ТТТ Protein: Met-Arg-Ser-His-Leu-Asn-Lys-Glu-Ala-Gly-Tyr-Phe DNA: ATG CGC TCC CAT CTT AAC CAA AGA GGT TGG CTA TTT T Protein: Met-Arg-Ser-His-Leu-Asn-Gln-Arg-Gly-Trp-Leu-Phe- missense nonsense antisense O frame-shift silentTranslate the following mRNA nucleotide sequence into an amino acid sequence, starting at the second base: 5’ - UGUCAUGCUCGUCUUGAAUCUUGUGAUGCUCGUUGGAUUAAUUGU - 3’The genetic disorder sickle-cell anemia occurs when the amino acid valine takes the place of glutamate during translation of a hemoglobin chain. Using the table of codons below, determine the mutation in DNA that produces this disorder. 1st position ✓ U C A G Select one: U C serine phenylalanine phenylalanine serine leucine serine leucine serine leucine leucine leucine leucine isoleucine isoleucine isoleucine methionine Table of mRNA Codons 2nd position valine valine valine valine proline proline proline proline alanine alaninc alanine alanine A tyrosine tyrosine a. CUC changes to C AG b. GAA changes to GUU c. CTT changes to CAT d. C A G changes to CTC stop stop threonine asparagine threonine asparagine threonine threonine histidine histidine arginine arginine glutamine arginine glutamine arginine lysine lysine G cysteine cysteine stop tryptophan aspartate aspartate glutamate glutamate serine serine arginine arginine glycine glycine glycine glycine 3rd position DCMO U С A G U C A G…