Essentials of Genetics (9th Edition) - Standalone book
9th Edition
ISBN: 9780134047799
Author: William S. Klug, Michael R. Cummings, Charlotte A. Spencer, Michael A. Palladino
Publisher: PEARSON
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Chapter 13, Problem 14PDQ
Review the concept of colinearity in Section 12.5 (p. 223) and consider the following question: Certain mutations called amber in bacteria and viruses result in premature termination of polypeptide chains during translation. Many amber mutations have been detected at different points along the gene that codes for a head protein in phage T4. How might this system be further investigated to demonstrate and support the concept of colinearity?
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According to the Central Dogma, genes are the blueprints for making proteins. Each gene (humans have 21,325) contains a single “coded message” of DNA bases (A, T, G, & C) attached in a specific order, which the cell “reads” to create an mRNA molecule that is then translated into protein. Knowing this, EXPLAIN how a SINGLE gene can make different proteins in different cells.
Consider the now dominant variant of the SARS-CoV-2 called the D614G mutation:
a) The mutation changes an Aspartate (D, Asp) to a Glycine (G, Gly) at nucleotide position 614 (that’s why it’s called the D614G mutant) in the S1 subunit of the Spike protein. Using only the information above and a codon table, what are the mRNA codon sequences of the 2019-dominant and 2020-dominant Spike proteins? Note the figure above is not needed toanswer the question. Report in 5’ to 3' orientation __________________________________________
b) What type of substitution is this? In your answer, address the following: • The expected substitution in base sequence (e.g., A à C)• If the mutation is synonymous, nonsynonymous, or a frameshift• If the mutation is a transition or a transversion
c) The mutation increases infectivity by reducing the stability of the Spike protein such that it can remain in the open conformation more often. The open conformation increases the chances of binding to the host ACE2…
A full-length eukaryotic gene is inserted into a bacterial chromosome. The gene contains a complete promoter sequence and a functional polyadenylation sequence, and it has wild-type nucleotides throughout the transcribed region. However, the gene fails to produce a functional protein.
a)List at least 3 possible reasons why this eukaryotic gene is not expressed in bacteria.
b)What changes would you recommend to permit expression of this eukaryotic gene in a bacterial cell?
Chapter 13 Solutions
Essentials of Genetics (9th Edition) - Standalone book
Ch. 13 - CASE STUDY | Crippled ribosomes Diamond Blackfan...Ch. 13 - CASE STUDY | Crippled ribosomes Diamond Blackfan...Ch. 13 - Prob. 3CSCh. 13 -
HOW DO WE KNOW?
1. In this chapter, we focused on...Ch. 13 - Review the Chapter Concepts list on p. 238. These...Ch. 13 - List and describe the role of all molecular...Ch. 13 - Contrast the roles of tRNA and mRNA during...Ch. 13 - Francis Crick proposed the adaptor hypothesis for...Ch. 13 -
6. During translation, what molecule bears the...Ch. 13 - Summarize the steps involved in charging tRNAs...
Ch. 13 - Each transfer RNA requires at least four specific...Ch. 13 -
9. Explain why the one-gene:one-enzyme hypothesis...Ch. 13 - Prob. 10PDQCh. 13 - Prob. 11PDQCh. 13 - Prob. 12PDQCh. 13 - Assuming that each nucleotide is 0.34 nm long in...Ch. 13 - Review the concept of colinearity in Section 12.5...Ch. 13 -
15. In your opinion, which of the four levels of...Ch. 13 -
16. List and describe the function of as many...Ch. 13 - How does an enzyme function? Why are enzymes...Ch. 13 -
18. Shown in the following table are several...Ch. 13 -
19. Three independently assorting genes are known...Ch. 13 -
20. How would the results in cross (a) of Problem...Ch. 13 - A series of mutations in the bacterium Salmonella...Ch. 13 -
22. The emergence of antibiotic-resistant strains...
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- 1. a)What would happen if the aminoacyl tRNA synthetase responsible for charging alanine tRNAs also charged methionine tRNAs with alanine? b)What would happen if an individual was homozygous for mutant alleles of the gene encoding the aminoacyl tRNA synthetase responsible for charging leucine tRNAs?arrow_forwardTransformation is a process in which bacteria take up new DNA released by dead cells and integrate it into their own genomes (see p. 265 in Chapter 9). In Streptococcus pneumoniae (which causes many cases of pneumonia, inner-ear infections, and meningitis), the ability to carry out transformation requires from 105 to 124 genes, collectively termed the com regulon. The com regulon is activated in response to a protein called competence-stimulating peptide (CSP), which is produced by the bacteria and exported into the surrounding medium. When enough CSP accumulates, it attaches to a receptor on the bacterial cell membrane, which then activates a regulator protein that stimulates the transcription of genes within the com regulon and sets in motion a series of reactions that ultimately result in transformation. Does the com regulon in Streptococcus pneumoniae exhibit positive or negative control? Explain your answer.arrow_forwardThe figure below shows the introns and exons found in gene X. The size of each exon and intron is shown as well. A study on this organism found that two mature mRNA molecules are produced for this gene. One is 457 nucleotides in length, and the other is 439 nucleotides in length. Name the process responsible for producing this variation. Also explain how these 457 and 439 nucleotide fragments were produced by referring to the information provided. Hint: This organism produces a poly-A tail of 120 nucleotides.arrow_forward
- The figure below shows the introns and exons found in gene X. The size of each exon and intron is shown as well. A study on this organism found that two mature mRNA molecules are produced for this gene. One is 457 nucleotides in length, and the other is 439 nucleotides in length. Name the process responsible for producing this variation. Also explain how these 457 and 439 nucleotide fragments were produced by referring to the information provided. Hint: This organism produces a poly-A tail of 120 nucleotides. 99 62 120 84 102 27 117 Gene X E1 11 E2 12 ЕЗ 13 E4 Exon (E) Intron (I)arrow_forwardThe figure below shows the introns and exons found in gene X. The size of each exon and intron is shown as well. A study on this organism found that two mature MRNA molecules are produced for this gene. One is 457 nucleotides in length, and the other is 439 nucleotides in length. Name the process responsible for producing this variation. Also explain how these 457 and 439 nucleotide fragments were produced by referring to the information provided. Hint: This organism produces a poly-A tail of 120 nucleotides. 99 62 120 84 102 27 117 Gene X E1 в в 11 E2 12 E4 Exon (E) Intron (1)arrow_forwardThe following gene sequence of nucleotides is found on the template (non-coding) strand of a molecule of DNA from a bacterial cell. The promoter of the gene is highlighted in bold letters and the +1 is underlined. Use the genetic code at the end of this packet to answer the following questions. 3'-AGGCATATTACGATGCCGGTACTTGATGATGACGGACCCATTATAGGACATATG-5' a) What is the sequence of the mRNA strand that will be transcribed from this piece of DNA? Indicate which is the 5’ and which is the 3’ end of the mRNA. b) What is the amino acid sequence that will be translated from this piece of DNAarrow_forward
- 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.arrow_forwarda) What is a mutation in molecular terms? b) a mutation deletes a base in the genomic DNA discuss how that will affect the reading frame and expression product production. Using the following list of codons describe, using diagrams etc., how information stored in the DNA is translated into a peptide. Be sure to discuss all steps. In other words, use a diagram and give me sequences, transcription and translation steps. Show the sequences of the sense and the other DNA strand, the mRNA and the tRNA’s. UUU -phenylalanine UCU -serine AUG –initiation/methionine CUU -leucine ACU -threonine GUU -valine UAA -Terminationarrow_forward(c) By binding one L-tryptophan molecule/monomer, the trp repressor binds to DNA to suppress syn- thesis of L-tryptophan in E. coli. Below is the amino acid sequence of the helix – (reverse) turn – helix region of the trp repressor that binds to DNA compared to the sequence of the corresponding DNA binding motif of the Prl protein, a different type of repressor protein. A diagram of the trp repressor dimer is also shown. reverse turn trp helix 4 70 Trp -Gly-Glu-Met-Ser-Gln-Arg-Glu-Leu-Lys-Asn-Glu-Leu-Gly-Ala-Gly- Ile- Prl -Ser-Glu-Glu-Ala-Lys-Glu-Glu-Leu-Ala-Lys-Lys-Cys-Gly-Ile-Thr- Val- Pri heilix trp helix 5 80 90 Trp Ala-Thr-Ile-Thr-Arg-Gly-Ser sgn-Ser-Leu-Lys-Ala-Ala- Prl Ser-Gln-Val-Ser-Asn-Trp-Phe-Gly-Asn-Lys-Arg-Ile-Arg- Prl helixarrow_forward
- Breast cancer can be caused by a genetic mutation on the BRCA1 gene changing a methionine to an arginine residue in the transcribed protein. How will this mutation effect this protein? a) Polarity before and after mutation: b) Size of the region before and after the mutation: c) Tertiary interaction you would expect with substrate: d) Name an amino acid that the unaffected protein's methionine could interact:arrow_forwardGive 7 examples where a specific nucleotide sequences/elements are recognized by protein or protein/RNA complex, from lecture 3.1-3.6. at least one examples from prokaryotic transcription, eukaryotic transcription, RNA processing, RNA stability, translation respectively. For each example, list the specific DNA/RNA sequences/elements and the protein or complex. E.g. eukaryotic promoter, transcription factor, transcription initiation.arrow_forwardHemophilia in the Russian royal family was caused by defective protein involved in blood clotting (factor IX). This defective protein was caused by a mutation that altered the splicing of the exons. This genetic change in the splicing pattern created a new stop codon in the mRNA for factor IX. What effect might this new stop codon have on the primary and tertiary levels of the mutant factor IX protein (compared to the native or wild-type protein).arrow_forward
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