Below is a segment found in the DNA template strand of a gene for a structural protein. This section of code will become an intron region of a gene. A mutation of the DNA causes the loss of the following CA nucleotides. What will the result be? 3' ATTCAGGGGACAGCGCCCCATTTA 5' O This mutation will result in the deletion of one amino acid during translation, changing the protein's ability to fold O This mutation will throw off the reading frame of the transcribed mRNA and the structural protein will not be made 4 This mutation is not important because the intron will be spliced out after transcription O This mutation will result in a substitution of one amino acid during translation, changing the protein's properties

Below is a segment found in the DNA template strand of a gene for a structural protein. This section of code will become an intron region of a gene. A mutation of the DNA causes the loss of the following CA nucleotides. What will the result be? 3' ATTCAGGGGACAGCGCCCCATTTA 5' O This mutation will result in the deletion of one amino acid during translation, changing the protein's ability to fold O This mutation will throw off the reading frame of the transcribed mRNA and the structural protein will not be made 4 This mutation is not important because the intron will be spliced out after transcription O This mutation will result in a substitution of one amino acid during translation, changing the protein's properties

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter9: Gene Expression And Gene Regulation

Section: Chapter Questions

Problem 16QP: Given the following tRNA anticodon sequence, derive the mRNA and the DNA template strand. Also,...

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Given the following mRNA, write the double-stranded DNA segment that served as the template. Indicate both the 5 and the 3 ends of both DNA strands. Also write out the tRNA anticodons and the amino acid sequence of the protein encoded by the mRNA message. DNA: mRNA: 5-CCGCAUGUUCAGUGGGCGUAAACACUGA-3 protein: tRNA:
Figure 28.41 gives some examples of recombination in IgG codons 95 and 96, as specified by the Vkand Jkgenes. List the codon possibilities and the amino acids encoded if recombination occurred in codon 97. Which of these possibilities is less desirable?
The 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’
3’ TACAATGGGCGACGCGCTTCGTTTCAGATT 5’ 5’ ATGTTACCCGCTGCGCGAAGCAAAGTCTAA 3’ Make vertical lines between codons to make this assignment easier to do. Which strand is the template strand? The first strand is the template strand as it going 3-5 Copy the template strand in mRNA. Label the 5’ and 3’ ends. 5' AUG UUA CCC GCU GCG CGA AGC AAA GUC UAA 3” Write out the amino acid (you can use the short form) that this protein would be made of (keep in mind proteins are usually a minimum of 50 proteins. Met-Leu-Pro-Ala-Ala-Arg-Ser-Lys-Val What do you notice about the mRNA strand compared to the non template strand of DNA? 2. What amino acid does the second codon code for on the DNA template strand? ______Assume the 6th amino acid in the strand is changed from a T to a C. What amino acid does it code for now? _______ What type of mutation is this? 3. Assume the 6th amino acid is changed from T to G on the DNA template strand. What type of mutation is this? What effect would…
Here is the nucleotide sequence of an imaginary strand of DNA: 5’ AATTGGCCATGC 3’. If this strand of DNA was transcribed, the resulting messenger mRNA molecule would be: Met-Val-Tyr-Lys 3’ TACGTACG 5’ 3’ TTAACCGGTACG 5’ 3’ UUAACCGGUACG 5’
A segment of a polypeptide chain is Arg-Gly-Ser-Phe-Val-Asp-Arg. It is encoded by the following segment of DNA: GGCTAGCTGCTTCCTTGGGGA CCGATCGACGAAGGAACCCCT Note out the mRNA sequence generated by the template strant to produce that polypeptide chain Label each stran with its correct polarity (5' and 3' ends on each strand)
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
A mRNA has the following sequence and direction: 5’-AUGUCAACCUAA-3’ What would be the anticodon sequences formed from this mRNA during the translation process?
Given the following DNA sequence: 3'-TACTTNGTNCTNTCN-5' where N stands for any nucleotide, give the complementary mRNA sequence. Indicate direction of strand as 3'--> 5' or 5'--> 3' as in the given sequence above. Give the amino acid sequence of your mRNA sequencelin No. 1. Indicate direction of strand as above. Use all lowercase letters, 3-letter name of amino acid separated by a hyphen (-), no spaces in-between.
If 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- G
An mRNA transcript has the following complete sequence: 5'-AUGCCUACGUUACGGACC-3' Rewrite the sequence to give it a missense mutation. Circle/highlight the mutation. Explain how your specific mutation would affect the formation and structure of the 2nd (middle) Base of a Codon 1st U CA G 3rd Base Base UUU Phe UUC Phe UUA Leu UUG Leu UCU Phe UCC Phe UCA Leu UCG Leu UAU Tyr UAC Tyr UAA STOP UAG STOP UGU Cys UGC Cys UGA STOP UGG Trp protein. U CUU Leu CUC Leu CUA Leu CUG Leu CCU Pro ССС Pro CCA Pro CCG Pro CAU His CAC His CAA Gln CAG Gln CGU Arg CGC Arg CGA Arg CGG Arg AGU Ser AGC Ser AGA Arg AGG Arg AAU Asn AAC Asn AUU lle AUC lle AUA lle AUG Met ACU Thr ACC Thr ACA Thr ACG Thr A AAA Lys AAG Lys GUU Val GUC Val GUA Val GUG Val GCU Ala GCC Ala GCA Ala GCG Ala GAU Asp GAC Asp GAA Glu GAG Glu GGU Gly GGC Gly GGA Gly GGG Gly UCAG
3’ TACAATGGGCGACGCGCTTCGTTTCAGATT 5’ 5’ ATGTTACCCGCTGCGCGAAGCAAAGTCTAA 3’ Copy the template strand in mRNA. Label the 5’ and 3’ ends. Write out the amino acid (you can use the short form) that this protein would be made of (keep in mind proteins are usually a minimum of 50 proteins.