Biology with Connect Access Card
10th Edition
ISBN: 9780077705701
Author: Raven, Peter
Publisher: McGraw-Hill Education
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Textbook Question
Chapter 15, Problem 3S
Describe how each of the following mutations will affect the final protein product (protein begins with start codon). Name the type of mutation.
Original template strand:
3′ – CGTTACCCGACCGTACGATTAGG–5′
3′ – CGTTACCCGAGCCGTAACGATTAGG –5′
3′ – CGTTACCCGATCCGTACGATTAGG –5′
3′ – CGTTACCCGAGCCGTTCGATTAGG –5′
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The sequence of the coding strand of a bacterial gene is given below. The positions of the first nine bases are numbered for your convenience. A missense mutation was introduced at position seven where the C was changed to a T resulting a mutant gene.
123456789
5'- ATGGCCCGACCGCAACTTTTCCGAGCTCTGGTGTCTGCGCAGTGACC-3
a. Write the template DNA (complementary strand) sequence for the wild type gene above
b. Write the DNA sequence of the mutant gene (Both DNA strands)
c. Write the sequence of mRNA produced from the mutant gene
d. Write the sequence of the mutant protein using the codon usage table provided in the end of this document.
Consider the following DNA template:
5’-AAGAGGTTCCAATGCAGGCACTCACCAACTCTTAAATAAA-3’
3’-TTCTCCAAGGTTACGTCCGTGAGTGGTTGAGAATTTATTT-5’
If the bottom DNA strand is used as template to transcribe mRNA, predict the amino acid sequence that would result from the process of translation.
Met-Ala-Leu-Thr-Gln-Glu-Gly
Met-Gly-Ser-Leu-Asn-Ser-Gln
Met-Thr-Asn-Ser-Leu-Ala-Gln
Met-Gln-Ala-Leu-Thr-Asn-Ser
Met-Glu-Ala-His-Trp-Ser-Tyr
2) Create an MRNA strand based on the given DNA template strand:
TACTTCCTATTITCTTGTCA CCGCACT
3) Using the mRNA codon chart, determine the amino acid sequence for the MRNA
sequence determined in question 3.
4) Consider the following double-stranded DNA molecule:
Complementary Strand: ATGTGTAGTGCGAGTTGA
Template Strand:
TACACATCACGCTCAACT
a) What would be the amino acid sequence coded for by the template strand of the
DNA molecule above?
Chapter 15 Solutions
Biology with Connect Access Card
Ch. 15 - The experiments with nutritional mutants in...Ch. 15 - What is the central dogma of molecular biology? a....Ch. 15 - In the genetic code, one codon a. consists of...Ch. 15 - Eukaryotic transcription differs from prokaryotic...Ch. 15 - An anticodon would be found on which of the...Ch. 15 - RNA polymerase binds to a ________ to initiate...Ch. 15 - During translation, the codon in mRNA is actually...Ch. 15 - You have mutants that all affect the same...Ch. 15 - The splicing process a. occurs in prokaryotes. b....Ch. 15 - The enzyme that forms peptide bonds is called...
Ch. 15 - In comparing gene expression in prokaryotes and...Ch. 15 - The codon CCA could be mutated to produce a. a...Ch. 15 - An inversion will a. necessarily cause a mutant...Ch. 15 - What is the relationship between mutations and...Ch. 15 - Prob. 1SCh. 15 - Frameshift mutations often result in truncated...Ch. 15 - Describe how each of the following mutations will...Ch. 15 - There are a number of features that are unique 10...
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biology and related others by exploring similar questions and additional content below.Similar questions
- What is the sequence of the mRNA transcript that will be produced from the following sequence of DNA? The top strand is the template strand, the bottom strand is the coding strand. 5’ – TCGGGATTAGACGCACGTTGGCATACCTCG – 3’ 3’ – AGCCCTAATCTGCGTGCAACCGTATGGAGC – 5’ Enter the mRNA sequence here (pay close attention to the direction of the molecule!): 5'-_____-3'arrow_forwardHelp me pleasearrow_forwardWhich of the following represents the sequence of an RNA transcript for which the coding strand (also known as non-template strand) of DNA has the sequence: GTACTGGCTAGCTGCTAGAA? Note all sequences are written 5'-3'. OA. AAGAUCGUCGAUCGGUCAUG OB. AAGATCGTCGATCGG TCATG OC. GTACTGGC TAGCTGC TAGAA OD. GUACUGGCUAGCUGCUAGAAarrow_forward
- The sequence of part of an mRNA transcript is What is the sequence of the DNA coding strand? 5'- 5' - AUGGGGAACAGCAAGAGUGGGGCCCUGUCCAAGGAG - 3' 5'- ATGGGGAACAGCAAGAGTGGGGCCCTGTCCAAGGAG What is the sequence of the DNA template strand? TACCCCTTGTCGTTCTCACCCCGGGACAGGTTCCTC Incorrect -3' -3'arrow_forwardBelow 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?arrow_forwardThe 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 Glyarrow_forward
- 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.arrow_forwardThe sequence below is of the DNA duplex for a gene in which transcription begins with the nucleotide highlighted by the arrow. If the upper strand shown is the template strand, write the sequence you expect for the mRNA transcribed from this gene. Please write 5' to 3'. 5'-[x]-3' 5'-TACGTGACGGTAATACTAGC-3' 3'-ATGCACTGCCATTATGATCG-5'arrow_forwardThe following are DNA fragments containing a small gene. The top strand is the coding strand. Transcribe all groups and translate. FIND THE POSSIBLE MUTATIONS Group B - MUTATION 5’-GGCAATGGGTTTGTGAAATTCTAAAAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACTTTAAGATTTTCAAAAATTAAG-5’ Group C- 5’-GGCAATGGGTTTGTGCAATTCTAAGAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACGTTAAGATTCTCAAAAATTAAG-5’arrow_forward
- Use a codon chart determine the amino acid sequence. Remember to read through the strand and ONLY start after the promoter and STOP when it tells you to stop. Follow example below: Example: DNA AGA TATA TAC CTC CGG TGG GTG CTT GTC TGT ATC CTT CTC AGT ATC MRNA O protein AUG GAG GCC ACC CAC GAA CAG ACA UAG GAA GAG UCA UAG start-glu-ala-thre-hist - asp-glu-threo-stop met DNA CCT ATA TAC ACA CGG AGG GTA CGC TAT TCT ATG ATT ACA CGG TTG CGA TCC ATA ATC mRNA DGGA UAU) AUG uGul Gcc nccl cAul GCol protein ly Tur MeT cys AlA ser HIJ Ala 2 3 4 DNA AGA ACT ATA TAC CTC TTA ACA CTC TAA AGA CCA GCA CTC CGA TGA ACT GGA GCA mRNA protein DNA TAT ATAC CTT GGG GAA TAT ACA CGC TGG CTT CGA TGA ATC CGT ACG GTA CTC GCC ATC mRNA protein D DNA TAA ACT ATA TAC CTA GCT TAG ATC TAA TTA CCC ATC mRNA protein Auu UGA UAU AGU GAUCGA AUC MAG Auu AAU leu Stop. TRY-Met-Asp- ARG-Isle-Stop-Ile. Asn DNA CTA TTT ATA TAC TAG AGC GAA TAG AAA CTT ATC ATC mRNA protein D DNA CAT ATA TAC CTT AGT TAT CCA TTG ACT CGA ATT GTG CGC TTG…arrow_forwardUse the genetic code table to determine the amino acid sequence of the given message strand of DNA below, from N to C terminal. All introns were removed in this sequence for simplicity. Write the amino acids in their 3-letter abbreviation and separate with a dash. Do not put a space in between characters so that LMS will recognize your answer. Message Strand +1 5-TAGTAGGCGGCATGTTTTCCCATACAGATGAAGGATAAACTCGTCT[x]TAT-3' [x]-cleavage site for CFI/CFII endonuclease (for RNA) Genetic Code: Second letter с A G UAUTyr UGC Cys UAC. UAA Stop UGA Stop UAG Stop UGG Trp CAUT CGU CAC His CGC CAA CGA Arg Gin CAGJ CGG AAU Asn AGU Ser AAC. AGC AAA AGA AAG Lys AGG Arg GAU GGU Asp GACJ GGC GAA GGA Glu GAGJ GGG] First letter כ A G U UUU UUC UUA UUGL CUU CUC CUA CUG AUU AUC lle AUA AUG Met GUU GUC Val GUA GUG Phe Leu Leu UCU UCC UCA UCG CCU CCC CCA CCG ACU ACC ACA ACG GCU GCC GCA GCG Ser Pro Thr Ala Gly DUAU DUAU DURO DURO A G Third letterarrow_forwardGiven the following DNA sequence of the template strand for a given gene: 5' TTTCCGTCTCAGGGCTGAAAATGTTTGCTCATCGAACGC3' Part A ) Write the mRNA that will be transcribed from the DNA sequence above (be sure to label the 5' and 3' ends). Part B ) Use the genetic code to write the peptide sequence translated in a cell from the mRNA in part A. Please use the 3 letter abbreviation for each amino acid. Part C: How would the peptide synthesized in a cell be different if the mRNA was translated in vitro (i.e. not in the cell)?arrow_forward
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