Concept explainers
Based on what you know about protein and DNA, why did Hershey and Chase pick 32P and 35S to label their phage?
To explain: The reason for picking phosphorus 32P and sulfur 35S by Hersey and Chase to label their phage.
Introduction: Every organism in this world contains genetic material which passes from one generation to another. The genetic material decides the physical appearance and other factors of an organism.
Explanation of Solution
Hersey and Chase experimented to identify the molecule which is responsible for heredity. They took advantage of the fact that protein present in protein coat of T2 virus contains sulfur but not phosphorus, whereas DNA present in genetic material of T2 virus contains phosphorus but not sulfur.
Therefore, Hersey and Chase prepared two batches of phage particles with various radioactive labeling. They used radioactive phosphorus 32P to label phage protein and sulfur 35S to label the protein of T2 virus. When 35S labeled T2 virus was allowed to infect bacteria, their progeny did not carry any radioactivity in their DNA. When bacteria were allowed to infect by 32P labeled T2 viruses, the radioactivity in the DNA of new phage particles was detected.
Hersey and Chase experiment showed that DNA acts as genetic material which passes from one generation to another, not protein.
Want to see more full solutions like this?
Chapter 16 Solutions
Becker's World of the Cell (9th Edition)
Additional Science Textbook Solutions
HUMAN ANATOMY
Biology: Concepts and Investigations
Campbell Biology (10th Edition)
Genetics: Analysis and Principles
Human Physiology: An Integrated Approach (8th Edition)
- Imagine that you are a student in Alfred Hershey and Martha Chase’s lab in the late 1940s. You are given five test tubes containing E. coli bacteria infected with T2 bacteriophages that have been labeled with either 32P or 35S. Unfortunately, you forget to mark the tubes and are now uncertain about which were labeled with 32P and which with 35S. You place the contents of each tube in a blender and turn it on for a few seconds to shear off the phage protein coats. You then centrifuge the contents to separate the protein coats and the cells. You check for the presence of radioactivity and obtain the following results. Which tubes contained E. coli infected with 32P-labeled phage? Explain your answer. Tube number Radioactivity present in 1 Cells 2 Protein coats 3 Protein coats 4 Cells 5 Cellsarrow_forwardImagine that you are a student in Alfred Hershey and Martha Chase’s lab in the late 1940s. You are given five test tubes containing E. coli bacteria infected with T2 bacteriophages that have been labeled with either 32P or 35S. Unfortunately, you forget to mark the tubes and are now uncertain about which were labeled with 32P and which with 35S. You place the contents of each tube in a blender and turn it on for a few seconds to shear off the phage protein coats. You then centrifuge the contents to separate the protein coats and the cells. You check for thepresence of radioactivity and obtain the following results. Which tubes contained E. coli infected with 32P-labeled phage? Explain your answer.arrow_forwardYou want to product human immunoglobulin G using a recombinant Escherichia coli system. Design the whole procedure and strategy by yourself.arrow_forward
- 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 experiment? * O Sulfur is part of the DNA molecule but not part of a protein molecule. Sulfur and phosphorous are some of the most reactive molecules and are easily traced. Sulfur and phosphorous are able to survive the centrifuge, a crucial component of the experiment. O Phosphorous is part of the DNA molecule but not part of a protein molecule.arrow_forwardIn the name E. coli 4X5B, which part is called the serotype, serovar, or strain? E coli 4X5Barrow_forwardE. coli cells are simultaneously infected with two strains of phage λ. One strain has a mutant host range, is temperature sensitive, and produces clear plaques (genotype h st c); another strain carries the wildtype alleles (genotype h+ st+ c+). Progeny phages are collected from the lysed cells and are plated on bacteria. The following numbers of different progeny phages are obtained: Progeny phage genotype Number of plaques h+ c+ st+ 321 h c st 338 h+ c st 26 h c+ st+ 30 h+ c st+ 106 h c+ st 110 h+ c+ st 5 h c st+ 6 a. Determine the order of the three genes on the phage chromosome. b. Determine the map distances between the genes. c. Determine the coefficient of coincidence and the interferencearrow_forward
- Phage T2 is estimated to consist of about 200,000 deoxyribonucleotides pairs. Give the length in micrometer of its DNA complement.arrow_forwardRefer to the following illustration to answer the question_ The illustration shows: O a lysogenic phage O a lytic phage new DNA O replicative transposition sequence non-replicative transposition O site-specific recombinationarrow_forwardWhy bacteriophage is called t4?arrow_forward
- One mL of a bacteriophage suspension is mixed with 20 mL of a bacterialculture and 50% of the phages adsorb. We know that the bacteriophagesuspension had a concentration of 1x10^10 viruses per mL, and the bacterialculture had a concentration of 3x10^8 bacteria per mL. What fraction of the cells is uninfected?arrow_forwardAre they true or false? a)If someone reads an absorbance of 0,25 at 260 nm, the concentration of the dsDNA is 25 ug/ml. b)If someone reads an absorbance of 0,25 at 260 nm, the concentration of the dsDNA is 25 ug/ml c)The copy number refers to the number of molecules of an individual plasmid that are normally found in a single bacterial cell. d)The amount of mRNA level can be determined via competitor RT-PCR in which two different sized products can be obtained, one of which is a known standard.arrow_forwardOne mL of a bacteriophage suspension is mixed with 20 mL of a bacterialculture and 50% of the phages adsorb. We know that the bacteriophagesuspension had a concentration of 1x10^10 viruses per mL, and the bacterialculture had a concentration of 3x10^8 bacteria per mL. What fraction of the cells is single infected?arrow_forward
- Human Anatomy & Physiology (11th Edition)BiologyISBN:9780134580999Author:Elaine N. Marieb, Katja N. HoehnPublisher:PEARSONBiology 2eBiologyISBN:9781947172517Author:Matthew Douglas, Jung Choi, Mary Ann ClarkPublisher:OpenStaxAnatomy & PhysiologyBiologyISBN:9781259398629Author:McKinley, Michael P., O'loughlin, Valerie Dean, Bidle, Theresa StouterPublisher:Mcgraw Hill Education,
- Molecular Biology of the Cell (Sixth Edition)BiologyISBN:9780815344322Author:Bruce Alberts, Alexander D. Johnson, Julian Lewis, David Morgan, Martin Raff, Keith Roberts, Peter WalterPublisher:W. W. Norton & CompanyLaboratory Manual For Human Anatomy & PhysiologyBiologyISBN:9781260159363Author:Martin, Terry R., Prentice-craver, CynthiaPublisher:McGraw-Hill Publishing Co.Inquiry Into Life (16th Edition)BiologyISBN:9781260231700Author:Sylvia S. Mader, Michael WindelspechtPublisher:McGraw Hill Education