1.6 Nucleic Acids · Educerie AP Biology

Educerie · AP · Biology

AP Biology 1.6 Nucleic Acids

In one breath

DNA and RNA are polymers of nucleotides, each a five-carbon sugar (deoxyribose in DNA, ribose in RNA), a phosphate and a nitrogenous base (A, T, G, C, or U in RNA).
Covalent bonds from the 3′ carbon of one sugar to the 5′ phosphate of the next make the backbone, so every strand has a 5′ phosphate end and a 3′ hydroxyl end, and new nucleotides are added only at the 3′ end.
The information is the base sequence.
DNA is two antiparallel strands in a double helix, held by hydrogen bonds between A–T and C–G pairs, so the strands are complementary and in double-stranded DNA A = T and G = C.
RNA is usually single-stranded and pairs A with U.

What you must be able to do

Describe the three parts of a nucleotide and the five basesLabel sugar, phosphate and base on a drawing; name the sugar in DNA and in RNA
Describe how nucleotides join into a strand with a 5′ end and a 3′ end, and that new nucleotides are added to the 3′ endFind the 3′ end on a diagram; say which end a new nucleotide joins
Describe DNA as an antiparallel double helix held by A–T and C–G pairs, and RNA's A–U pairingWrite the partner of a given strand with its ends labelled; calculate base percentages
Describe the structural differences between DNA and RNASugar, base and number of strands, as a comparison

Once you have read the note, tick off each one you can do in Check your understanding.

In the exam. Section I multiple choice: labelling the parts of a nucleotide, finding the 5′ and 3′ ends of a strand in a diagram, writing or choosing the complementary strand, telling DNA from RNA by sugar, base or number of strands, and data on base composition (percentages of A, T, G and C). Unit 6 (replication, transcription) leans on all of it. In Section II it supplies points in Analyze model or visual representation (describe a drawing of a DNA segment), Conceptual analysis (explain why the strands must be antiparallel or why base pairing matters) and Analyze data (a table of base percentages).

1The nucleotide: sugar, phosphate, base

Every nucleotide has the same three parts, joined by covalent bonds:

A five-carbon sugar. In DNA it is deoxyribose; in RNA it is ribose. Deoxyribose has one fewer oxygen atom ("de-oxy"), which is all the difference between them.
A phosphate group, which contains phosphorus and is negatively charged.
A nitrogenous base, a ring-shaped molecule containing nitrogen. DNA uses A, T, G and C; RNA uses A, U, G and C.

Only the base differs from one nucleotide to the next in a strand, so only the bases carry information. The CED does not examine the molecular structure of individual nucleotides: you need the three parts and the five base names, not ring drawings.

2A strand has direction: 5′ and 3′

The carbons of the sugar are numbered 1′ to 5′ ("one prime" to "five prime"; the prime separates sugar carbons from atoms in the base). Two of those numbers name the ends of every strand. Figure 1 walks through it.

Figure 1 · The 5′ and 3′ ends of a strand, and an antiparallel partner

Number the sugar's carbons. The base is attached to carbon 1′. The phosphate is attached to carbon 5′, which sticks out of the ring.

Each link runs from carbon 3′ to the next phosphate. A covalent bond joins the 3′ carbon of one sugar to the phosphate on the 5′ carbon of the next. Repeated, these links make the sugar–phosphate backbone, with the bases sticking out to one side.

The two ends are different. At the top, the first nucleotide's phosphate is free: the 5′ end. At the bottom, the last sugar's 3′ carbon carries a free hydroxyl group (–OH): the 3′ end. A strand is read and written in the direction 5′ to 3′ (the arrow).

The partner strand runs the other way. In DNA a second strand pairs with the first, base to base, A with T and C with G, held by hydrogen bonds (the dashed lines). Its 5′ end is at the bottom and its 3′ end at the top: the two strands are antiparallel.

When a cell builds a nucleic acid, each new nucleotide joins by its phosphate to the free –OH at the 3′ end of the growing strand, by a dehydration synthesis that makes a new covalent bond. So strands only ever grow at their 3′ end, in the 5′ to 3′ direction. This one fact explains a great deal of Unit 6, where you meet the enzymes that do it.

Nucleotides are only ever added to the 3′ end, so every strand grows 5′ to 3′, and the two strands of DNA run in opposite directions.

3DNA: an antiparallel double helix

A DNA molecule is two strands of nucleotides, with their backbones on the outside and their bases pointing inward, paired across the middle. The pairing is fixed:

Adenine pairs with thymine (A–T).
Cytosine pairs with guanine (C–G).

The pairs are held by hydrogen bonds, which are weak one by one, so the strands can be separated when DNA is copied or read, but strong in their thousands, so the molecule is stable. A C–G pair is held by three hydrogen bonds and an A–T pair by two, as the dashes in Figure 1 show. The two strands twist around each other into a double helix.

Because each base has only one partner, the strands are complementary: knowing one strand tells you the other exactly. That is how DNA can be copied, and why the order of bases on one strand is enough to store information.

Worked example: writing a complementary strand. A strand reads 5′-ATGCCA-3′. Pair each base, and remember the partner runs the other way:

given strand: 5′- A T G C C A -3′

partner strand: 3′- T A C G G T -5′

partner, written 5′ to 3′: 5′-TGGCAT-3′

Both lines for the partner are the same molecule. Exam answers often expect sequences written 5′ to 3′, so the safest habit is to label the ends every time.

4Base percentages: a worked example

In double-stranded DNA every A is paired with a T and every G with a C. So the amount of A equals the amount of T, and the amount of G equals the amount of C.

An invented sample of double-stranded DNA from a fungus is 21% adenine. What are the other percentages?

T = A = 21%

A + T = 42%, so G + C = 100 − 42 = 58%

G = C = 58 ÷ 2 = 29%

Check: 21 + 21 + 29 + 29 = 100. The rule holds only for double-stranded DNA. In single-stranded RNA, or the single-stranded DNA of some viruses, A need not equal T (or U) at all. In a data table, an organism whose A and T percentages differ widely almost certainly has a single-stranded genome. That is evidence, not a rule broken.

5RNA, and how it differs from DNA

RNA is built the same way, nucleotides joined 5′ to 3′ by covalent bonds, but differs in three structural ways that the CED names:

Feature	DNA	RNA
Sugar	deoxyribose	ribose
Base that pairs with adenine	thymine (T)	uracil (U)
Strands	usually double-stranded	usually single-stranded

When RNA does pair with another strand, or with DNA, adenine pairs with uracil (A–U) and cytosine still pairs with guanine. In an exam, write each difference as a comparison ("DNA contains deoxyribose, whereas RNA contains ribose"), not as two separate facts.

Practice questions

5 questions on this note, from easy to hard. Each one you get right lights a star of this topic's constellation.