Theme A: A1 Molecules - A1.2 Nucleic Acids

Question 1

DNA

Answer 1

stands for deoxyribonucleic acid. the molecule that provides the long-term stored genetic information for all organisms in Earth.

Question 2

structure of a nucleotide (single-stranded - RNA)

Answer 2

nucleotides are made up of a phosphate group, a pentose sugar, and a nitrogenous base. nucleotides form a chain/polymer as a result of condensation reactions that form covalent bonds between the sugar of 1 nucleotide and the phosphate group of the next. this is also known as sugar-phosphate bonding that forms a strong backbone in the molecule.

Question 3

nitrogenous bases

Answer 3

4 bases in RNA: guanine, adenine, cytosine, uracil.
4 bases in DNA: guanine, adenine, cytosine, thymine.
The sequence of nitrogenous bases are genes or genetic messages that mainly code for amino acids, which are molecules that combine to form proteins.

Question 4

condensation reaction

Answer 4

the reaction releases a water molecule when synthesising DNA or RNA molecules.

Question 5

complementary bases

Answer 5

adenine (A) always bonds with thymine (T).
cytosine (C) always bonds with guanine (G).

Question 6

structure of DNA

Answer 6

has a double helix shape made of 2 antiparallel strands of nucleotides. the strands are linked by hydrogen bonds between complementary bases.

Question 7

similarities between DNA and RNA

Answer 7

they’re both linear polymers consisting of sugars, phosphates, and bases.

Question 8

difference between DNA and RNA

Answer 8

DNA: double stranded
RNA: single stranded

DNA: deoxyribose sugar
RNA: ribose sugar

DNA: thymine is 1 of the 4 bases
RNA: uracil is 1 of the four bases

DNA: double helix shape
RNA: variety of shapes depending on type of RNA

DNA: permanent genetic code of a cell/organism
RNA: only RNA viruses contain permanent genetic code

Question 9

difference between deoxyribose and ribose sugars

Answer 9

ribose has 1 more oxygen atom which can be exemplified through their molecular formulas:
deoxyribose: C5H10O4
ribose: C5H10O5

Question 10

examples of nucleic acids

Answer 10

1) messenger RNA (mRNA)
2) transfer RNA (tRNA)
3) ribosomal RNA (rRNA)
4) adenosine triphosphate (ATP)

Question 11

why is complementary base pairing useful?

Answer 11

it plays an essential role in DNA replication. when a protein called helicase breaks the hydrogen bonds between the DNA into 2 singe strands, these free-floating individual nucleotides pair with unmatched nucleotides to make an exact copy of the original molecule.

Question 12

triplet codon (“triplet”)

Answer 12

every 3 bases represent a meaningful piece of genetic information called a triplet or triple codon.

Question 13

directionality of RNA and DNA strands

Answer 13

strands of DNA and RNA are synthesised starting with the 5’ (5 prime) end and working towards the 3’ (3 prime) end. the 5’ end of a single strand is the unbound phosphate group and the 3’ end terminates with a sugar.

Question 14

purine-to-pyrimidine bonding

Answer 14

purine is always bonded to a pyrimidine. therefore, the 2 DNA strands maintain the same distance from one another with 3 rings in total, leading to a stable double helix shape.
purines (double ring size and thus larger): adenine and guanine
pyrimidines (single ring size and thus smaller): thymine and cytosine

Question 15

structure of a nucleosome

Answer 15

the DNA wraps twice around the 8 histone proteins at the nucleosome’s core. 1 additional histone is used to maintain its shape. linker DNA then connects it to the next nucleosome

Question 16

efficient packaging solution of DNA

Answer 16

DNA, which is very long, wraps around 8 histone proteins to make nucleosomes that are connected via linker DNA. nucleosomes coil around other proteins into a condensed shape known as a chromosome. A human cell contains 46 chromosomes in its nucleus.

Question 17

resources and techniques used in the Hershey-Chase experiment

Answer 17

Alfred Hershey and Martha Chase used a bacteriophage (a virus that infects bacteria; it’s composed of a protein outer coats and inner core of DNA) to infect the bacterium E.coli. they also used the radioisotopes phosphorus 32 (found in DNA) and sulfur 35 (found in 2 of the 20 amino acids present in protein).

Question 18

conclusion of the Hershey-Chase experiment

Answer 18

The E.coli infected by the bacteriophage with phosphorus had radioactivity detected inside of the cell, meaning the DNA was passed from the virus to bacteria. Whereas the one infected by the phage with sulfur in the protein coat had no radioactivity detected inside of it, meaning the protein was not passed on. This allowed them to conclude that DNA, not protein, is the genetical material.

Question 19

techniques that lead to Chargaff’s rule

Answer 19

Erwin Chargaff developed paper chromatography to show the proportion of nitrogenous bases in various sources of DNA.

Question 20

discoveries from Chargaff’s experiment

Answer 20

Chargaff’s experiment revealed that the proportion of adenine (A) is equal to thymine (T), and the proportion of guanine (G) is equal to cytosine (C), a finding now known as Chargaff’s rules. This discovery helped to falsify the tetranucleotide theory, which suggested that DNA consisted of a repetitive sequence of the four nucleotide bases in equal amounts (A, T, G, and C) and was therefore too simple to carry genetic information.