The technique of Western blotting is described in Chapter 21 and also in experimental question E5. Let’s suppose a researcher is interested in the effects of mutations on the expression of a protein-encoding gene that encodes a protein we will call protein X. This protein is expressed in skin cells and contains 572 amino acids. Its molecular mass is approximately
Lane 1: A normal individual
Lane 2: An individual who is homozygous for a deletion that removes the promoter for this gene
Lane 3: An individual who is heterozygous in which one gene is normal and the other gene has a mutation that introduces an early stop codon at codon 421 Lane 4: An individual who is homozygous for a mutation that introduces an early stop codon at codon 421
Lane 5: An individual who is homozygous for a mutation that changes codon 198 from a valine codon into a leucine codon
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Genetics: Analysis and Principles
- In Northern blot analysis, mRNA samples from tissues are bound to a labeled DNA probe that is complementary to the mRNA, and run on a gel to be visualized. The protein tropomyosin is known to be present in both brain and liver. When brain and liver tissue were assayed for the presence of tropomyosin mature mRNA, bands of two different sizes were seen. Tropomyosin gene diagram (3000 bp total): Shown in attatched image If the band on the Northern blot for mRNA isolated from liver tissue is 2580 bp, whereas from brain tissue the band is 2250 bp, what is most likely? a)The two mRNAs are made from different tropomyosin DNA sequences. b)Exon 2 is alternatively spliced out of the brain mRNA. c)Introns 1 and 2 are spliced out of the brain transcript but not the liver transcript. d)Exons 1 and 3 are spliced out of the brain transcript but not the liver transcript. e)Exon 2 is alternatively spliced out of the liver mRNA.arrow_forwardIn the Western blot shown here, proteins were isolated from redblood cells and muscle cells from two different individuals. Oneindividual was unaffected, and the other suffered from a diseaseknown as thalassemia, which involves a defect in hemoglobin. Theblot was exposed to an antibody that recognizes β globin, whichis one of the polypeptides that constitute hemoglobin. Equal totalamounts of cellular proteins were added to each lane. Explain these results.arrow_forwardIn this western blot, the levels of phosphorylated TBK (PTBK) decrease with increasing amounts/expression of the viral protein (VP). Figure description: Increasing amounts of a plasmid expressing the viral protein (0.5, 1, or 2ug) were cotransfected with TBK1 expression plasmid. Cells were harvested 24 h post-transfection and analyzed for phosphorylated TBK1 (anti- TBK1 Ser172), total TBK1 (anti-TBK1), B-actin (anti-B-actin), and viral protein (anti-VP) expression by Western blot analysis. VP PTBK1 (S172) TBK1 actin Virus proteins O True O Falsearrow_forward
- antibodies were produced in a rabbit against a 25 KDa human soluble protien. when the rabbit antiserum was used in a western blot of human soluble protiens, it detected 25-KDa protien,but also bound to protiens of 65KDa and 39 KDa.assuming that a pure protien was used to make the antibodies,how do yuo explain the results of the western blot?arrow_forwardA purified recombinant protein is analyzed for molecular weight by SDS-PAGE at pH 8.5. From the protein sequence deduced from the gene that was expressed in bacteria, the protein is expected to have a molecular weight of 44,000. However, the molecular weight of the protein is found by SDS-PAGE to be 52,000. Explain the reason or reasons for this difference in molecular weight. What calculation could you make to help explain this discrepancy?arrow_forwardAnalyzing Cloned Sequences A base change (A to T) is the mutational event that created the mutant sickle cell anemia allele of beta globin. This mutation destroys an MstII restriction site normally present in the beta globin gene. This difference between the normal allele and the mutant allele can be detected with Southern blotting. Using a labeled beta globin gene as a probe, what differences would you expect to see for a Southern blot of the normal beta globin gene and the mutant sickle cell gene?arrow_forward
- What is the purpose of Southern's blotting technique? Explain in detail the biochemical principle that underpin each step of the method.arrow_forwardIn this western blot, the levels of TBK remain constant with increasing amounts/expression of the viral protein (VP). Figure description: Increasing amounts of a plasmid expressing the viral protein (0.5, 1, or 2ug) were cotransfected with TBK1 expression plasmid. Cells were harvested 24 h post-transfection and analyzed for phosphorylated TBK1 (anti-pTBK1 Ser172), total TBK1 (anti-TBK1), B-actin (anti-B-actin), and viral protein (anti-VP) expression by Western blot analysis. VP PTBK1 (S172) TBK1 actin Virus proteins O True O Falsearrow_forwardThe following table lists 4 bacterial strains that are partial diploids for lac operon genes. Given the activity of beta-galactosidase measured for each strain in the absence (-lac) or presence (+lac) of lactose, complete the table by choosing the appropriate symbol (+, -, C, S) to indicate the allele of the gene or site missing from the table (blue numbers). strain A BC 5 C D 7 chromosome I O 1 2 4 1 [Select] 9 3 [Select] [Select] [Select] 9 [Select] + + Z + + 6 + I +5 + 10 plasmid O 3 + 7 C Z + 8 8 2 [Select] 4 [Select] 6 [Select] B-gal act. -lac +lac 0.002 0.003 0.002 0.058 0.063 0.121 0.059 0.062 Select] 1 ✔ [ Select] + is C Sarrow_forward
- Materials In order to determine the genetic material of a T2 phage, Alfred Hershey and Martha Chase conducted experiments using T2 phages that infected bacteria. In one treatment, they grew phages with radioactive sulfur. In another treatment, they grew phages with radioactive phosphorous. They allowed both types of phages to infect bacterial cells. After infection, they found that only bacteria infected with phages grown with radioactive phosphorous showed any radioactivity. Why did they use radioactive sulfur and phosphorous for this Updates Grades Members O Conferences DBQ Online experiment? * Newsela ormation O Sulfur is part of the DNA molecule but not part of a protein molecule. Biology Periods 1 and 2 Sulfur and phosphorous are some of the most reactive molecules and are easily ding periods school MP1, Highschool Highschool MP3, school MP4 traced. Sulfur and phosphorous are able to survive the centrifuge, a crucial component of the experiment. ion Phosphorous is part of the DNA…arrow_forwardA Western blot is seen in the picture above. Sodium dodecyl sulfate gel polyacrylamide gel electrophoresis was used to isolate the proteins (SDS-PAGE).1. Since gel electrophoresis has isolated individual proteins, what kind of molecule is used to detect them? 2. On the right side of this Western blot, three molecules are identified. One of the bands corresponds to the smallest molecule?arrow_forwardMass spectrometry is a powerful tool in proteomics. What are the four key features of a mass spectrometer? Describe briefly how MALDI and two-dimensional polyacrylamide gel electrophoresis could be used to identify a protein expressed in cancer cells but not in normal healthy cells.arrow_forward
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning