Molecular Biology Exam quizzes and
answers graded A+
Describe the structural differences between a base, a nucleoside, and a nucleotide. –
ANS✅✅Nitrogenous bases include thymine, cytosine, uracil (RNA), adenine, and guanine. These
bases are like building blocks to a nucleoside and nucleotide. A nucleoside is a connected sugar and
base, while a nucleotide is a connected sugar, base, and phosphate group. A nucleoside can have a
pyrimidine or purine base, which differs in where the connection of the base and pentose is located.
A nucleotide has a phosphate group connected to the 5' carbon of the sugar, which creates a
phosphoester bond.
Adenine and guanine are double-ringed structures but differ at C2 and C6. Guanine has an amine
group at C2, while adenine lacks an amine group at C2. Guanine has a keto group at C6, while
adenine has an amine group at C6. Thymine and cytosine are single-ringed structures but differ at C4
and C5. Thymine has a keto group at C4, while cytosine has an amine group at C4. Thymine has a
methyl group at C5, while cytosine lacks the methyl group at C5.
HW 2
Avery, MacLeod, and McCarty concluded that DNA contained the genetic information to transform
nonpathogenic R (rough) Streptococcus pneumoniae to pathogenic S (smooth) S. pneumoniae.
Explain the experimental logic behind treatment of the active purified fraction with
deoxyribonuclease, ribonuclease, or proteolytic enzymes. - ANS✅✅The experimental logic with
treating the active purified fraction with deoxyribonuclease, ribonuclease, and proteolytic enzymes
was to show that DNA was the transforming material in the newly formed S cells.
Utilizing S and R bacteria (Virulance was known to depend on a polysaccharide capsule that protects
the bacterium, S bacteria = virulent, R bacteria = non-virulent). Heat killed S cells were purified so
that only genetic material remained. Treated with a series of -ases. When treated with
deoxribonuclease and added to live R cell - DNA was destroyed and no transformation occurred.
When treated with ribonuclease and added to live R cells - RNA was destroyed and transformation
occurred. When treated with proteases and added to live R cells - protein was destroyed and
transformation occurred. Demonstrated that the transforming principle is not protein or RNA and
has to be DNA.
HW 2
Considering Chargaff's data, explain how the human data help refute that thymine pairs with
cytosine in double stranded DNA. - ANS✅✅Chargaff's rules explained that purines always pair with
pyrimidines based on the molar content of each base. The base composition is expressed as a
fraction à Thymine and cytosine are both pyrimidines, therefore, would not pair with each other.
If a human genome has 24% thymine and 26% cytosine, there are not enough thymines to base pair
with cytosines. However, that same genome would have 24% thymine, 24% adenine, 26% cytosine
and 26% guanine (because chargaffs data demonstrated that A=T and G=C), therefore in order to
, have each base paired up within the double helix A has to base pair with T and G has to base pair
with C.
HW 2
What are the 8 key features of the Watson-Crick Model for B-DNA. - ANS✅✅1) Two
polydeoxyribonucleotide strands twist about each other to form a double helix structure.
2) Phosphate and deoxyribose groups form the backbone on the outside of the double helix.
3) Purine and pyrimidine base pairs are stacked inside the helix, which forms planes that are
perpendicular to the helix axis.
4) The helix diameter is 2.0 nm or 20 angstroms.
5) Base pairs are separated by an average distance of 0.34 nm or 3.4 angstroms, and the structure
repeats itself, or turns, after about 10 base pairs or every 3.4 nm (34 angstroms).
6) Adenine always pairs with thymine, and cytosine always pairs with guanine (purine paired with
pyrimidine).
7) Two strands are antiparallel to one another, which means 5'- 3'/ 3'- 5'.
8) There is presence of a major and minor groove, where the minor groove is about 1.2 nm (12
angstroms) and the major groove is about 2.2 nm (22 angstroms).
HW 2
Write out the steps in the Central Dogma. List where each step occurs in the cell. For each arrow,
name the process that the arrow represents and the primary enzyme responsible for that step.
Name the steps in which mRNA, tRNA, and rRNA have a role and briefly describe the role of each in
the steps you name. - ANS✅✅The central dogma is DNA à RNA à protein.
DNA is transcribed into RNA.
RNA is translated into protein using mRNA, tRNA, and rRNA.
DNA is replicated by DNA replication.
RNA is reversed transcribed into cDNA, which is complementary DNA.
tRNA serves as a link between mRNA and the growing chain of amino acids that are in a protein.
rRNA reads the order of amino acids and links the amino acids together.
mRNA carries protein information from the DNA in the cell's nucleus to the cell's cytoplasm to be
coded for amino acids.
HW 2
Describe how proteins recognize specific base sequences within the DNA. Describe the recognition
pattern for each of the 4 bases. - ANS✅✅The specific base sequences are recognized from the
answers graded A+
Describe the structural differences between a base, a nucleoside, and a nucleotide. –
ANS✅✅Nitrogenous bases include thymine, cytosine, uracil (RNA), adenine, and guanine. These
bases are like building blocks to a nucleoside and nucleotide. A nucleoside is a connected sugar and
base, while a nucleotide is a connected sugar, base, and phosphate group. A nucleoside can have a
pyrimidine or purine base, which differs in where the connection of the base and pentose is located.
A nucleotide has a phosphate group connected to the 5' carbon of the sugar, which creates a
phosphoester bond.
Adenine and guanine are double-ringed structures but differ at C2 and C6. Guanine has an amine
group at C2, while adenine lacks an amine group at C2. Guanine has a keto group at C6, while
adenine has an amine group at C6. Thymine and cytosine are single-ringed structures but differ at C4
and C5. Thymine has a keto group at C4, while cytosine has an amine group at C4. Thymine has a
methyl group at C5, while cytosine lacks the methyl group at C5.
HW 2
Avery, MacLeod, and McCarty concluded that DNA contained the genetic information to transform
nonpathogenic R (rough) Streptococcus pneumoniae to pathogenic S (smooth) S. pneumoniae.
Explain the experimental logic behind treatment of the active purified fraction with
deoxyribonuclease, ribonuclease, or proteolytic enzymes. - ANS✅✅The experimental logic with
treating the active purified fraction with deoxyribonuclease, ribonuclease, and proteolytic enzymes
was to show that DNA was the transforming material in the newly formed S cells.
Utilizing S and R bacteria (Virulance was known to depend on a polysaccharide capsule that protects
the bacterium, S bacteria = virulent, R bacteria = non-virulent). Heat killed S cells were purified so
that only genetic material remained. Treated with a series of -ases. When treated with
deoxribonuclease and added to live R cell - DNA was destroyed and no transformation occurred.
When treated with ribonuclease and added to live R cells - RNA was destroyed and transformation
occurred. When treated with proteases and added to live R cells - protein was destroyed and
transformation occurred. Demonstrated that the transforming principle is not protein or RNA and
has to be DNA.
HW 2
Considering Chargaff's data, explain how the human data help refute that thymine pairs with
cytosine in double stranded DNA. - ANS✅✅Chargaff's rules explained that purines always pair with
pyrimidines based on the molar content of each base. The base composition is expressed as a
fraction à Thymine and cytosine are both pyrimidines, therefore, would not pair with each other.
If a human genome has 24% thymine and 26% cytosine, there are not enough thymines to base pair
with cytosines. However, that same genome would have 24% thymine, 24% adenine, 26% cytosine
and 26% guanine (because chargaffs data demonstrated that A=T and G=C), therefore in order to
, have each base paired up within the double helix A has to base pair with T and G has to base pair
with C.
HW 2
What are the 8 key features of the Watson-Crick Model for B-DNA. - ANS✅✅1) Two
polydeoxyribonucleotide strands twist about each other to form a double helix structure.
2) Phosphate and deoxyribose groups form the backbone on the outside of the double helix.
3) Purine and pyrimidine base pairs are stacked inside the helix, which forms planes that are
perpendicular to the helix axis.
4) The helix diameter is 2.0 nm or 20 angstroms.
5) Base pairs are separated by an average distance of 0.34 nm or 3.4 angstroms, and the structure
repeats itself, or turns, after about 10 base pairs or every 3.4 nm (34 angstroms).
6) Adenine always pairs with thymine, and cytosine always pairs with guanine (purine paired with
pyrimidine).
7) Two strands are antiparallel to one another, which means 5'- 3'/ 3'- 5'.
8) There is presence of a major and minor groove, where the minor groove is about 1.2 nm (12
angstroms) and the major groove is about 2.2 nm (22 angstroms).
HW 2
Write out the steps in the Central Dogma. List where each step occurs in the cell. For each arrow,
name the process that the arrow represents and the primary enzyme responsible for that step.
Name the steps in which mRNA, tRNA, and rRNA have a role and briefly describe the role of each in
the steps you name. - ANS✅✅The central dogma is DNA à RNA à protein.
DNA is transcribed into RNA.
RNA is translated into protein using mRNA, tRNA, and rRNA.
DNA is replicated by DNA replication.
RNA is reversed transcribed into cDNA, which is complementary DNA.
tRNA serves as a link between mRNA and the growing chain of amino acids that are in a protein.
rRNA reads the order of amino acids and links the amino acids together.
mRNA carries protein information from the DNA in the cell's nucleus to the cell's cytoplasm to be
coded for amino acids.
HW 2
Describe how proteins recognize specific base sequences within the DNA. Describe the recognition
pattern for each of the 4 bases. - ANS✅✅The specific base sequences are recognized from the
