Human Heredity: Principles and Issues (MindTap Course List)
11th Edition
ISBN: 9781305251052
Author: Michael Cummings
Publisher: Cengage Learning
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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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Here is part of a gene:
GTAACCGTATTGCAGCTATTAGCAGCCATG CATTGGCATAACGTCGATAATCGTCGGTAC
If the bottom strand of the DNA carries the gene, write the mRNA that would be transcribed from the bottom strand of the DNA:
Use 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.
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 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 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_forward
- 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 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_forwardIn 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_forward
- 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 Glyarrow_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_forwardTranscribe the following strand in Mrna, then translate it into amino acidsarrow_forward
- Use 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_forwardThe following DNA fragment was isolated from the beginning of a gene. Determine which strand is transcribed, indicate the polarity of the two DNA strands, and then give the sequence of bases in the resulting mRNA and its polarity. CCCTACGCCTTTCAGGTT GGGATGCGGAAAGTCCAAarrow_forwardIf DNA segments changes from GCATAG to GCATA, this is a: MRNA Codon/Amino Acid Chart First Base Second Base Third Base U A G 0001 Phenylalanine UCU UAU1 Tyrosine (Tyr) UAC UGUT FCcysteine (Cys) UGCJ U UUCJ (Phe) UCC Serine (Ser) UCA U UUA1 UAAT UGA - Stop A FLeucine (Leu) UUG- FStop UAG- UCG- UGG - Tryptophan (Trp) G CU- CCU CGUT CAU1 Histidine (His) CAC U CUC FLeucine (Leu) CUA CC Proline (Pro) CCA CGC FArginine (Arg) CGA CAA1 Glutamine A CAGI (Glu) CGG- CUG- CCG- G AUU AAU1 Asparagine ACU1 AGUT FSerine (Ser) AGC- AUC FIsoleucine (le) ACC Threonir AACJ (Asn) A AUA- ACA (Thr) AAA1 FLysine (Lys) AAG- AGA, FArginine (Arg) AGG- A Start Methionine (Met) ACG- AUG - GUU- GCU GAU- GGU | Aspartic Acid GAÇJ(Asp) U GỤC Valine (Val) GUA GCC FAlanine (Ala) GGC Glycine (Gly) GGA G GAA1 Glutamic Acid A GCA GCG- GAGJ (Glu) GGG- GUG- Garrow_forward
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