A1.2 Nucleic Acids

Question 1

DNA and RNA full names (A1.2.1)

Answer 1

Deoxyribonucleic acid
Ribonucleic acid

Question 2

Nucleic acids (A1.2.1)

Answer 2

Very large molecules, made from subunits called nucleotides which link to form a polymer.

RNA + DNA

Question 3

Genetic material(/hereditary information) (A1.2.1)

Answer 3

A store of information. If copied, can be passed from cell to cell and also from parent to offspring.

DNA is the genetic material of all living organisms.

Question 4

Components of a nucleotide (A1.2.2)

Answer 4

Pentose sugar - 5 carbons
Phosphate group - acidic + neg. charged part
Base - Contains nitrogen

Question 5

Sugar–phosphate “backbone” (A1.2.3)

Answer 5

Links nucleotides together into chain + conserves base sequence -> covalent bonds between phosphate group of one nucleotide and pentose of the next.

Question 6

Bases in each nucleic acid that form the basis of a code (A1.2.4)

Answer 6

Sequences of bases is how genetic information is stored. Info stored in a coded form—universal genetic code shared by all organisms.

Question 7

Nucleic acids as polymers formed by condensation of nucleotide monomers (A1.2.5)

Answer 7

OH of phosphorus + pentose used. -> One group combined w/ H from the other, producing H2O.

Remaining O forms new covalent bond, linking nucleotides.

Question 8

DNA structure (A1.2.6)

Answer 8

Double helix made of two antiparallel strands of nucleotides, linked by hydrogen bonding between complementary base pairs.

Question 9

Why is it important that DNA strands run antiparallel (A1.2.6)

Answer 9

One strand ends with the phosphate group of the terminal nucleotide while the other strand ends with a deoxyribose. This allows bases to form hydrogen bonds.

Question 10

Differences between DNA and RNA (A1.2.7)

Answer 10

DNA double stranded, RNA single stranded.

Uracil is present instead of thymine in RNA.

Pentose sugar DNA = deoxyribose. RNA = ribose. Deoxyribose -> one fewer oxygen (no O on C2).

Question 11

Role of complementary base pairing in allowing genetic information to be replicated (A1.2.8)

Answer 11

Strands separate. Original strands = templates.

Newly synthesized strand has same base sequence as other template strand.

Semi-conservative replication: one DNA molecule -> two identical DNA molecule.

Question 12

Role of complementary base pairing in allowing genetic information to be expressed (A1.2.8)

Answer 12

A gene contains info needed for a particular purpose. The info having effect on cell is known as gene expression.

The first stage of gene expression is transcribing (copying) the base sequence using one template strand.

Question 13

Conservation of the genetic code across all life forms as evidence of universal common ancestry (A1.2.10)

Answer 13

Groups of three bases = codons -> have meanings in the code. 64 different codons. Genetic code universal w/ minor exceptions.

Most codons code particular amino acid. 1 signals protein synthesis start, 3 signal stop.

Question 14

Directionality of DNA and RNA (A1.2.11)

Answer 14

Pentose of nucleotide at one end is unlinked= 3′ terminal. C3 is available for linkage to another nucleotide.

Phosphate group of nucleotide at other end is unlinked = 5′ terminal (P group attached to C5 of PS).

Question 15

Directionality of DNA on replication (A1.2.11)

Answer 15

5′ to 3′ -> nucleotides added to 3′ end of growing polymer. 5′ phosphate of free nucleotide links to sugar at 3′ end of growing polymer.

One strand, 5′ to 3′ assembly moves same direction as overall process of replication. The other it moves opposite.

Question 16

Directionality of DNA and RNA on transcription (A1.2.11)

Answer 16

5′ to 3′ -> nucleotides added to the 3′ end of the growing polymer.

One DNA strand is template -> allows the assembly of RNA strand to move in same direction as overall transcription process.

Question 17

Directionality of RNA on translation (A1.2.11)

Answer 17

5′ to 3′.
rRNA that carries out translation moves along the RNA molecule towards the 3′ end.

The RNA molecule carries sequence information for polypeptide synthesis which occurs by linking amino acids together.

Question 18

Purine-to-pyrimidine bonding (A1.2.12)

Answer 18

Adenine + guanine = purine bases (2 rings of atoms)

Cytosine + thymine = pyrimidine (1 ring)

Question 19

Purine-to-pyrimidine bonding as a component of DNA helix stability (A1.2.12)

Answer 19

Each base pair has 1 purine and pyrimidine.

Two base pairs are equally width + require same distance between the two backbones -> helps make the structure of DNA stable + allows any sequence of bases in genes.

Question 20

Structure of a nucleosome (A1.2.13)

Answer 20

Nucleosome core = 8 histone proteins (2x4) makes a disc-shaped structure.

DNA wound approx. 2x around core + short linker DNA between adjacent nucleo.

H1 reinforces binding + can help package chromos. when nucleus prepairs to divide.

Question 21

Evidence from the Hershey–Chase experiment for DNA as the genetic material (A1.2.14)

Answer 21

Viruses contained proteins with radioactive 35S and others that contained DNA with radioactive 32P.

Used blender to separate the non-genetic component of the virus. Then centrifuge -> bacteriophages + bacteria separate through density -> concentrates cells in a pellet.

Measure radioactivity in the pellet and the supernatant.