Human Heredity: Principles and Issues (MindTap Course List)
11th Edition
ISBN: 9781305251052
Author: Michael Cummings
Publisher: Cengage Learning
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Textbook Question
Chapter 9, Problem 8QP
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. Draw the primary transcript and the mRNA resulting from this DNA.
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The following double stranded segment of DNA is part of a protein coding gene. The segments in uppercase letters (ACTG) represent the exons. The segments in lowercase letters (acgt) represent introns. The lower strand is the template strand that is used by the RNA polymerase to make an RNA transcript. Draw or write-out a) the sequence of the primary transcript and b) the mature mRNA resulting from this stretch of DNA.
The following bacterial DNA sequence uses the top strand as the coding strand and the bottom strand as the template strand. Write what the messenger RNA sequence for this gene would be after transcription occurs.
The following sequence is the coding strand of a piece of DNA. Type out the corresponding
template strand of DNA and the MRNA that could be made from this piece of DNA if you assume
that transcription begins at the first nucleotide listed. Next, use the attached genetic code table to
translate the mRNA into a polypeptide that could be made from the message and list that
sequence along with those of the DNA and MRNA.
Coding strand
5'-TACCGTATGATTCTCTTGTATGGGTAACC-3'
Second letter
U
UUU
UUC
UCU
UAU
UGU
Phe
Tyr
Cys
UCC
UCA
UAC
UAA STOP
UAG STOP UGG| Trp
UGC
Ser
UUA
UGA STOP A
UUG Leu
UCG
CUU
CCU
CC
CAU
CGU
CGC
His
CUC
C
CUA
CAC
САА
Leu
Pro
Arg
CCA
CGA
Gln
CUG
СCG
CAG
CGG
AGU Ser
AAU
AAC
ACU
AUU
lle
AUC
Asn
ACC
ACA
AGC
Thr
AAA
AGA
AUA
AUG Met
Arg
ACG
AAG Lys
AGG
GAU
GCU
GCC
GUU
GGU
Asp
GAC
GGC
GGA
GUC
Val
Ala
Gly
GCA
GCG
GAA
GAG
GUA
Glu
GUG
GGG
Third letter
UCAG
UCAG
5CAG
UCAG
First letter
Chapter 9 Solutions
Human Heredity: Principles and Issues (MindTap Course List)
Ch. 9.6 - Antibiotics and Protein Synthesis Antibiotics are...Ch. 9.6 - Antibiotics and Protein Synthesis Antibiotics are...Ch. 9 - There have been recurring cases of mad-cow disease...Ch. 9 - There have been recurring cases of mad-cow disease...Ch. 9 - There have been recurring cases of mad-cow disease...Ch. 9 - The Link Between Genes and proteins The genetic...Ch. 9 - Define replication, transcription, and...Ch. 9 - If the genetic code used 4 bases at a time, how...Ch. 9 - If the genetic code uses triplets, how many...Ch. 9 - What is the start codon? What are the stop codons?...
Ch. 9 - Is an entire chromosome made into an mRNA during...Ch. 9 - The promoter and terminator regions of genes are...Ch. 9 - The following segment of DNA codes for a protein....Ch. 9 - What are the three modifications made to pre-mRNA...Ch. 9 - The pre-mRNA transcript and protein made by...Ch. 9 - Briefly describe the function of the following in...Ch. 9 - Prob. 12QPCh. 9 - Determine the percent of the following gene that...Ch. 9 - How many kilobases of the DNA strand below will...Ch. 9 - Prob. 15QPCh. 9 - Given the following tRNA anticodon sequence,...Ch. 9 - Given the following mRNA, write the...Ch. 9 - The following is a portion of a protein:...Ch. 9 - Below is the structure of glycine. Draw a...Ch. 9 - Indicate in which category, transcription or...Ch. 9 - Prob. 21QPCh. 9 - Polypeptide folding is often mediated by other...Ch. 9 - Do mutations in DNA alter proteins all the time?Ch. 9 - a. Can a mutation change a proteins tertiary...
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- A certain mRNA strand has the following nucleotide sequence: 5AUGACGUAUAACUUU3 What is the anticodon for each codon? What is the amino acid sequence of the polypeptide? (Use Figure 13-5 to help answer this question.) Figure 13-5 The genetic code The genetic code specifies all possible combinations of the three bases that compose codons in mRNA. Of the 64 possible codons, 61 specify amino acids (see Figure 3-17 for an explanation of abbreviations). The codon AUG specifies the amino acid methionine and also signals the ribosome to initiate translation (start). Three codonsUAA, UGA, and UAGdo not specify amino acids; they terminate polypeptide synthesis (stop).arrow_forwardThe following segment of DNA codes a protein. The uppercase letters represent Exons, the lowercase letters introns. Draw the pre- mRNA, the mature mRNA and translate the codons using the genetic code to form the protein. Identify the 5’UTR and 3UTR 5’- AGGAAATGAAATGCCAgaattgccggatgacGGTCAGCaatcgaGCACATTTGTGATTTACCGT-3’arrow_forwardThe sequence of coding strand of a DNA molecule is given below. Assume that it is read from left to right. CCTACCTTATGCCAAGTTGGGGATAAACTC The left end of this molecule is the end. How many amino acids will be in the protein translated from this sequence? What is the name (not abbreviation) of the fourth amino acid in the protein translated from this sequence? The label on the end of the protein that is translated first is the + Jend.arrow_forward
- Write the mRNA sequence (5′ to 3′) that would result from transcription of the DNA sequence. (Note: Ignore the colors)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 template (antisense) strand and the DNA molecule is transcribed, what is the correct nucleotide sequence and direction of the RNA formed after transcription?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_forward
- In the following table, below each DNA nucleotide, type in the complementary mRNA nucleotides. Then, for each set of three DNA and complementary mRNA nucleotides, use the amino acid chart to translate the nucleotides into amino acids, and type them below. DNA T-A-C A-A-G A-T-G G-G-G A-T-T mRNA Enter Text — Enter Text — Enter Text Enter Text — Enter Text — Enter Text Enter Text — Enter Text — Enter Text Enter Text — Enter Text — Enter Text Enter Text — Enter Text — Enter Text Amino acid Enter Text Enter Text Enter Text Enter Text Enter Textarrow_forwardThe following is a prokaryotic DNA sequence. Fill in the other strand of DNA. Be sure to label the ends appropriately (5’ and 3’). Transcription goes from left to right and stops after the right most base pair. On your DNA, indicate which is the coding/partner and which one is the template. Write out the pre-mRNA transcript. Be sure to label the ends appropriately (5’ and 3’). DNA 5’ – T G G G G G A T A C C G C A T G C A G G T A G T C T A A G C G A G T G A C – 3’ DNA mRNAarrow_forwardBelow is the 5’–3’ strand of a double-stranded DNA molecule with the following nucleotidesequences (all belong to an exon):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 template (antisense) strand and the DNA molecule is transcribed that produced a functional mRNA. Assuming there are no mutations, the said mRNA is then brought to the site of protein synthesis,a. what would be the amino acid sequence of the synthesized polypeptide chain?b. how many possible kinds of tRNA molecule that will bring the 2nd amino acid observing wobble hypothesis? List down their anticodons.arrow_forward
- The following eukaryotic DNA sequence is a made up gene. It is a mutated variant from the one that we used last week. The mutations can be seen underlined in purple (You can use your mature RNA to skip some steps, just make sure to change the nucleotides that are marked below. With this information please answer the questions below. What is the sequence of the polypeptide that will be generated from the mature mRNA? write a "-" between each amino acid. All in caps. Write STOP for the stop codon. Example ALA-LEU-MET-ILE...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_forwardGive the mRNA and amino acid sequence from this DNA strand: CCATTAACCTTACTGCTGGCTAAATTCGTTGCTarrow_forward
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