Bio 340 Module 5 And 6 Questions And Answers
With Verified Solutions Graded A+ Latest Update
2025.
Features of Hereditary Material - ANSWER 1. Localized to the nucleus and a component of
chromosomes
2. Able to accurately replicate itself so that daughter cells contain the same information as parent cells
3. Store information
4. Express information
Frederick Griffith Experiment: bacterial transformation - ANSWER Led to the discovery of
DNA by injecting mice with an with heat-killed S bacteria (that is, S bacteria that had been heated to high
temperatures, causing the cells to die). Unsurprisingly, the heat-killed S bacteria did not cause disease in
mice.
The experiments took an unexpected turn, however, when harmless R bacteria were combined with
harmless heat-killed S bacteria and injected into a mouse. Not only did the mouse develop pneumonia and
die, but when Griffith took a blood sample from the dead mouse, he found that it contained living S
bacteria.
Griffith concluded that the R-strain bacteria must have taken up what he called a "transforming principle"
from the heat-killed S bacteria, which allowed them to "transform" into smooth-coated bacteria and
become virulent.
Avery, MacLeod, McCarty - ANSWER Several lines of evidence suggested to Avery and his
colleagues that the transforming principle might be DNA:
1. The purified substance gave a negative result in chemical tests known to detect proteins, but a strongly
positive result in a chemical test known to detect DNA.
2. The elemental composition of the purified transforming principle closely resembled DNA in its ratio of
nitrogen and phosphorous.
3. Protein- and RNA-degrading enzymes had little effect on the transforming principle, but enzymes able
to degrade DNA eliminated the transforming activity.
These results all pointed to DNA as the likely transforming principle. However, Avery was cautious in
interpreting his results. He realized that it was still possible that some contaminating substance present in
small amounts, not DNA, was the actual transforming principle.
Hershey-Chase Experiment - ANSWER To establish whether the phage injected DNA or
protein into host bacteria, Hershey and Chase prepared two different batches of phage. In each batch, the
, phage were produced in the presence of a specific radioactive element, which was incorporated into the
macromolecules (DNA and protein) that made up the phage.
- One sample was produced in the presence of 35S, end text, a radioactive isotope of sulfur. Sulfur is
found in many proteins and is absent from DNA, so only phage proteins were radioactively labeled by
this treatment.
-The other sample was produced in the presence of 32P, a radioactive isotope of phosphorus. Phosphorus
is found in DNA and not in proteins, so only phage DNA (and not phage proteins) was radioactively
labeled by this treatment.
When Hershey and Chase measured radioactivity in the pellet and supernatant from both of their
experiments, they found that a large amount of 32P appeared in the pellet, whereas almost all of the 35S
appeared in the supernatant. Based on this and similar experiments, Hershey and Chase concluded that
DNA, not protein, was injected into host cells and made up the genetic material of the phage.
nucleotides (the building blocks of DNA) - ANSWER The important components of the
nucleotide are a nitrogenous base, deoxyribose (5-carbon sugar), and a phosphate group. The nitrogenous
base can be a purine such as adenine (A) and guanine (G), or a pyrimidine such as cytosine (C) and
thymine (T).
nucleotides combine.. - ANSWER by covalent bonds known as phosphodiester bonds or
linkages
deoxynucleoside monophosphates (dNMPs) - ANSWER DNA nucleotides that are part of a
polynucleotide chain have one phosphate residue attached to the hydroxyl group of the 5' carbon of one
sugar of one nucleotide and the hydroxyl group of the 3' carbon of the sugar of the next nucleotide,
thereby forming a 5'-3' phosphodiester bond
deoxynucleotide triphosphates (dNTPs). - ANSWER not part of a polynucleotide chain carry a
string of three phosphate groups at the 5' carbon and are identified as dATP, dGTP, dCTP, and dTTP
Base pairing takes place between.. - ANSWER ..a purine and pyrimidine; namely, A pairs with
T and G pairs with C. A and T form two H bonds, G and C form three.
Hydrogen bonds - ANSWER non-covalent bonds that form between the partial charges that are
associated with the hydrogen, oxygen, and nitrogen atoms of nucleotide bases
antiparallel - ANSWER the 3' end of one strand faces the 5' end of the other strand.
-example: one strand is 5'- ATCG -3', then the complementary strand is 3'- TAGC -5'.
Chargaff's rules state - ANSWER DNA from any cell of all organisms should have a 1:1 ratio
(base pair rule) of pyrimidine and purine bases and, more specifically, that the amount of G = C and the
amount of A = T.
The haploid human genome is 3 Gbp long (3x10^9 bp). How long would the genomic DNA in a single
diploid cell in your body be if it were stretched out end to end?
(hint: Remember that 1 Angstrom = 0.1 nm, and that each base pair is separated from the other base pair
by a distance of 3.4 Angstroms.) - ANSWER ~2 m
With Verified Solutions Graded A+ Latest Update
2025.
Features of Hereditary Material - ANSWER 1. Localized to the nucleus and a component of
chromosomes
2. Able to accurately replicate itself so that daughter cells contain the same information as parent cells
3. Store information
4. Express information
Frederick Griffith Experiment: bacterial transformation - ANSWER Led to the discovery of
DNA by injecting mice with an with heat-killed S bacteria (that is, S bacteria that had been heated to high
temperatures, causing the cells to die). Unsurprisingly, the heat-killed S bacteria did not cause disease in
mice.
The experiments took an unexpected turn, however, when harmless R bacteria were combined with
harmless heat-killed S bacteria and injected into a mouse. Not only did the mouse develop pneumonia and
die, but when Griffith took a blood sample from the dead mouse, he found that it contained living S
bacteria.
Griffith concluded that the R-strain bacteria must have taken up what he called a "transforming principle"
from the heat-killed S bacteria, which allowed them to "transform" into smooth-coated bacteria and
become virulent.
Avery, MacLeod, McCarty - ANSWER Several lines of evidence suggested to Avery and his
colleagues that the transforming principle might be DNA:
1. The purified substance gave a negative result in chemical tests known to detect proteins, but a strongly
positive result in a chemical test known to detect DNA.
2. The elemental composition of the purified transforming principle closely resembled DNA in its ratio of
nitrogen and phosphorous.
3. Protein- and RNA-degrading enzymes had little effect on the transforming principle, but enzymes able
to degrade DNA eliminated the transforming activity.
These results all pointed to DNA as the likely transforming principle. However, Avery was cautious in
interpreting his results. He realized that it was still possible that some contaminating substance present in
small amounts, not DNA, was the actual transforming principle.
Hershey-Chase Experiment - ANSWER To establish whether the phage injected DNA or
protein into host bacteria, Hershey and Chase prepared two different batches of phage. In each batch, the
, phage were produced in the presence of a specific radioactive element, which was incorporated into the
macromolecules (DNA and protein) that made up the phage.
- One sample was produced in the presence of 35S, end text, a radioactive isotope of sulfur. Sulfur is
found in many proteins and is absent from DNA, so only phage proteins were radioactively labeled by
this treatment.
-The other sample was produced in the presence of 32P, a radioactive isotope of phosphorus. Phosphorus
is found in DNA and not in proteins, so only phage DNA (and not phage proteins) was radioactively
labeled by this treatment.
When Hershey and Chase measured radioactivity in the pellet and supernatant from both of their
experiments, they found that a large amount of 32P appeared in the pellet, whereas almost all of the 35S
appeared in the supernatant. Based on this and similar experiments, Hershey and Chase concluded that
DNA, not protein, was injected into host cells and made up the genetic material of the phage.
nucleotides (the building blocks of DNA) - ANSWER The important components of the
nucleotide are a nitrogenous base, deoxyribose (5-carbon sugar), and a phosphate group. The nitrogenous
base can be a purine such as adenine (A) and guanine (G), or a pyrimidine such as cytosine (C) and
thymine (T).
nucleotides combine.. - ANSWER by covalent bonds known as phosphodiester bonds or
linkages
deoxynucleoside monophosphates (dNMPs) - ANSWER DNA nucleotides that are part of a
polynucleotide chain have one phosphate residue attached to the hydroxyl group of the 5' carbon of one
sugar of one nucleotide and the hydroxyl group of the 3' carbon of the sugar of the next nucleotide,
thereby forming a 5'-3' phosphodiester bond
deoxynucleotide triphosphates (dNTPs). - ANSWER not part of a polynucleotide chain carry a
string of three phosphate groups at the 5' carbon and are identified as dATP, dGTP, dCTP, and dTTP
Base pairing takes place between.. - ANSWER ..a purine and pyrimidine; namely, A pairs with
T and G pairs with C. A and T form two H bonds, G and C form three.
Hydrogen bonds - ANSWER non-covalent bonds that form between the partial charges that are
associated with the hydrogen, oxygen, and nitrogen atoms of nucleotide bases
antiparallel - ANSWER the 3' end of one strand faces the 5' end of the other strand.
-example: one strand is 5'- ATCG -3', then the complementary strand is 3'- TAGC -5'.
Chargaff's rules state - ANSWER DNA from any cell of all organisms should have a 1:1 ratio
(base pair rule) of pyrimidine and purine bases and, more specifically, that the amount of G = C and the
amount of A = T.
The haploid human genome is 3 Gbp long (3x10^9 bp). How long would the genomic DNA in a single
diploid cell in your body be if it were stretched out end to end?
(hint: Remember that 1 Angstrom = 0.1 nm, and that each base pair is separated from the other base pair
by a distance of 3.4 Angstroms.) - ANSWER ~2 m
