1. Researchers discovered that a single amino acid change of cysteine to tryptophan in a transmembrane...
1. Researchers discovered that a single amino acid change of cysteine to tryptophan in a transmembrane protein causes retinal degeneration.
1a. Using the genetic code, what mutant mRNA likely encodes this substitution and what is the WT mRNA sequence? (Indicate polarity)
1b. Is the substitution caused by a transition or transversion mutation?
Homework Answers
1.a. mRNA sequence UGU encode Cystein and UGG encode Tryptophan.
Wild-type mRNA: 5′--------UGU-------------3′
Mutant mRNA: 5′--------UGG-------------3′
(polarity of mRNA means the direction of 5' and 3').
1.b. The given substitution is caused by transversion mutation. In transversion mutation, a purine (two ring) is substituted for a pyrimidine (one ring) or vice versa. Here, purine Guanine is substituted for pyrimidine Uracil.
Add Answer to:
1. Researchers discovered that a single amino acid change of
cysteine to tryptophan in a transmembrane...
Below is the genomic DNA of gene X, a 3 exon gene that encodes a 131 amino acid single pass trans...
why is E the answer
Below is the genomic DNA of gene X, a 3 exon gene that encodes a 131 amino acid single pass transmembrane protein. Shown are the transcriptional start site, splice donor, acceptor and branch sites and translational start and stop codons. Transcriptional start EXON 1 INTRON 1 EXON 2 INTRON 2 EXON 3 Spfice Donor Splice Acceptor Polyadenylation signal Branch point 17. Treatment with ethidium bromide, an intercalating agent, caused DNA polymerase to add an extra...
Shown below are the amino acid sequences of the wild-type and three mutant forms of a...
Shown below are the amino acid sequences of the wild-type and three mutant forms of a short protein. Each mutation results from a single nucleotide change (transition/transversion / insertion / deletion). Use this information to answer the following questions. Hint: First, reconstruct as much as you can of the wild-type RNA sequence and then reference that sequence when analyzing the mutations. Wild type: met - gin-ala - ser-val - arg - phe Mutant 1: met - gln - pro-ser -...
Shown below are the amino acid sequences of the wild-type and three mutant forms of a...
Shown below are the amino acid sequences of the wild-type and three mutant forms of a short protein. Each mutation results from a single nucleotide change (transition / transversion / insertion / deletion). Use this information to answer the following questions. Hint: First, reconstruct as much as you can of the wild-type RNA sequence and then reference that sequence when analyzing the mutations. Wild type: met – gln – ala – ser – val – arg – phe Mutant 1:...
1. If an mRNA carried a mutation that caused a difference in the amino acid sequence...
1. If an mRNA carried a mutation that caused a difference in the amino acid sequence relative to the wild-type gene product, would the translation machinery recognize this and correct it? Explain. 2. Synthesis of mRNA (transcription) and protein (translation) proceed in an ordered manner, from one end of the molecule to another. What is the polarity of each of these processes? (Use the terms 5', 3', amino, carboxyl.)
You have a small gene that encodes the following amino acid: N-MET-ASP-SER-VAL-ALA-ARG-PHE-MET-TRP-C. There is a single...
You have a small gene that encodes the following amino acid: N-MET-ASP-SER-VAL-ALA-ARG-PHE-MET-TRP-C. There is a single mutation in the DNA that causes a change in the amino acid sequence to: N-MET-VAL-GLN-TRP-PRO-ASP-LEU-CYS-GLY-C. a) What kind of mutation is this? Explain. (2 points) b) Indicate the DNA sequence (coding strand) of the gene. Show the original DNA sequence then the mutated sequence. Wild type DNA: Mutant DNA: You have another mutation (a different mutation from the one described in parts a and...
Please note that Questions 15 to 17 are connected questions. Question 15: The following shows a...
Please note that Questions 15 to 17 are connected
questions.
Question 15:
The following shows a partial DNA sequence from the wild-type
(normal) allele for the human leukemia-linked apoptotic
gene.
5' ATGCGATTAATCGGTAAA 3' (non-template strand)
3' TACGCTAATTAGCCATTT 5' (template strand)
Please answer the following questions:
(a) If the bottom strand serves as the DNA template for
transcription, what is the resulting mRNA sequence?
The mRNA sequence is 5' 3'. (2
marks)
5' AUG CGA UUA AUC GGU AAA 3' ?
Please enter...
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genetic biology 5'-GCATGAGTCTGGTACGCTTTTAAAGC-3' 3-CATGCG-5' IIIII 3. (a) in the sequence above, what enzyme would you need...
genetic biology
5'-GCATGAGTCTGGTACGCTTTTAAAGC-3' 3-CATGCG-5' IIIII 3. (a) in the sequence above, what enzyme would you need to extend the short stretch of nucleotides shown on the bottom strand? (b) Write the sequence of the newly synthesized fragment and label its S' and 3' end. (c) The covalent bond between these adjacent nucleotides is what type of chemical bond? After using a chemical mutagen to generate mutations in a DNA sequence, scientists noted a mutation from C to T at the...
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why is E the answer
Below is the genomic DNA of gene X, a 3 exon gene that encodes a 131 amino acid single pass transmembrane protein. Shown are the transcriptional start site, splice donor, acceptor and branch sites and translational start and stop codons. Transcriptional start EXON 1 INTRON 1 EXON 2 INTRON 2 EXON 3 Spfice Donor Splice Acceptor Polyadenylation signal Branch point 17. Treatment with ethidium bromide, an intercalating agent, caused DNA polymerase to add an extra...
Shown below are the amino acid sequences of the wild-type and three mutant forms of a short protein. Each mutation results from a single nucleotide change (transition/transversion / insertion / deletion). Use this information to answer the following questions. Hint: First, reconstruct as much as you can of the wild-type RNA sequence and then reference that sequence when analyzing the mutations. Wild type: met - gin-ala - ser-val - arg - phe Mutant 1: met - gln - pro-ser -...
Please note that Questions 15 to 17 are connected
questions.
Question 15:
The following shows a partial DNA sequence from the wild-type
(normal) allele for the human leukemia-linked apoptotic
gene.
5' ATGCGATTAATCGGTAAA 3' (non-template strand)
3' TACGCTAATTAGCCATTT 5' (template strand)
Please answer the following questions:
(a) If the bottom strand serves as the DNA template for
transcription, what is the resulting mRNA sequence?
The mRNA sequence is 5' 3'. (2
marks)
5' AUG CGA UUA AUC GGU AAA 3' ?
Please enter...
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5'-GCATGAGTCTGGTACGCTTTTAAAGC-3' 3-CATGCG-5' IIIII 3. (a) in the sequence above, what enzyme would you need to extend the short stretch of nucleotides shown on the bottom strand? (b) Write the sequence of the newly synthesized fragment and label its S' and 3' end. (c) The covalent bond between these adjacent nucleotides is what type of chemical bond? After using a chemical mutagen to generate mutations in a DNA sequence, scientists noted a mutation from C to T at the...
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