A1.2 Nucleic Acids | IB Biology HL

studied byStudied by 70 people
5.0(2)
get a hint
hint

State the two primary functions of nucleic acids.

A1.2.1— DNA as the genetic material of all living organisms.

1 / 47

Tags and Description

48 Terms

1

State the two primary functions of nucleic acids.

A1.2.1— DNA as the genetic material of all living organisms.

1. Pass genetic information between generations.

2. Code for protein production.

New cards
2

State the two types of nucleic acids used in cells.

A1.2.1— DNA as the genetic material of all living organisms.

DNA: deoxyribonucleic acid

RNA: ribonucleic acid

<p>DNA: deoxyribonucleic acid</p><p>RNA: ribonucleic acid</p>
New cards
3

Outline the meaning and implication of DNA being the genetic material of all living organisms.

A1.2.1— DNA as the genetic material of all living organisms.

Meaning: All living organisms use DNA as the genetic material

Implication: The use of the genetic code across all forms of life is evidence of universal common ancestry of life. The sequences of DNA in cells can be analyzed and compared to determine evolutionary relationships between organisms. The more similar the sequence, the more closely related the organisms.

New cards
4

State why RNA viruses do not falsify the claim that all living things use DNA as the genetic material.

A1.2.1— DNA as the genetic material of all living organisms.

Some viruses use RNA as their genetic material. However, because viruses are not made of cells, they are not considered to be living.

New cards
5

List the three components of a nucleotide.

A1.2.2— Components of a nucleotide.

A nucleotide is the monomer subunit of the nucleic acids. A nucleotide has three component parts:

1. a nitrogenous base

2. A 5-carbon "pentose”sugar (ribose or deoxyribose)

3. A phosphate group

<p>A nucleotide is the monomer subunit of the nucleic acids. A nucleotide has three component parts:</p><p>1. a nitrogenous base</p><p>2. A 5-carbon &quot;pentose”sugar (ribose or deoxyribose)</p><p>3. A phosphate group</p>
New cards
6

Identify and label the carbons of a pentose sugar.

A1.2.2— Components of a nucleotide.

The carbons of the sugar component of the nucleotide are numbers clockwise, starting from the oxygen in the ring at the top and the phosphate group to the left.

<p>The carbons of the sugar component of the nucleotide are numbers clockwise, starting from the oxygen in the ring at the top and the phosphate group  to the left.</p>
New cards
7

Define “backbone” as related to nucleic acid structure.

A1.2.3— Sugar–phosphate bonding and the sugar–phosphate “backbone” of DNA and RNA.

The “backbone” is the alternating phosphate-sugar… found in a polymer of nucleic acids.

The relative strength of the backbone maintains the nucleotides in their specific sequence.

<p>The  “backbone” is the alternating phosphate-sugar… found in a polymer of nucleic acids.</p><p>The relative strength of the backbone maintains the nucleotides in their specific sequence.</p>
New cards
8

Explain how nucleotides can connect to form a nucleic acid polymer.

A1.2.3— Sugar–phosphate bonding and the sugar–phosphate “backbone” of DNA and RNA.

Nucleotides connect by creating covalent bonds between the sugar of one nucleotide and the phosphate group of another nucleotide in a condensation reaction.

The 5’ phosphate group on one nucleotide forms a new covalent bond with the 3' carbon on the pentose of the next nucleotide. Water is created as a byproduct.

<p>Nucleotides connect by creating covalent bonds between the sugar of one nucleotide  and the phosphate group of another nucleotide in a condensation reaction.</p><p>The 5’ phosphate group on one nucleotide forms a new  covalent bond with the 3&apos; carbon on the pentose of the next nucleotide.   Water is created as a byproduct.</p>
New cards
9

State the names of the nitrogenous bases found in DNA and RNA.

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

Cytosine (DNA and RNA)

Guanine (DNA and RNA)

Adenine (DNA and RNA)

Thymine (DNA only)

Uracil (RNA only)

<p>Cytosine (DNA and RNA)</p><p>Guanine (DNA and RNA)</p><p>Adenine (DNA and RNA)</p><p>Thymine (DNA only)</p><p>Uracil (RNA only)</p>
New cards
10

Outline how the sequence of bases in a nucleic acid serves as a ‘code.’

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

A code is a system in which one symbol signifies the meaning of another symbol.

In the genetic code, a group of three nucleic acid bases signifies for an amino acid. A gene is a specific sequence of DNA nucleotides that codes for the making of a specific protein.

New cards
11

Define gene.

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

A gene is a specific sequence of nitrogenous bases in DNA nucleotides that codes for the making of a protein.

New cards
12

Describe the condensation reaction that forms a polymer of RNA from RNA nucleotides.

A1.2.5— RNA as a polymer formed by condensation of nucleotide monomers.

RNA nucleotides connect by creating covalent bonds between the ribose sugar of one nucleotide and the phosphate group of another nucleotide in a condensation reaction.

The 5’ phosphate group on one RNA nucleotide forms a new covalent bond with the 3' carbon on the ribose of the next nucleotide. Water is created as a byproduct.

<p>RNA nucleotides connect by creating covalent bonds between the ribose sugar of one nucleotide and the phosphate group of another nucleotide in a condensation reaction.<br><br>The 5’ phosphate group on one RNA nucleotide forms a new covalent bond with the 3' carbon on the ribose of the next nucleotide. Water is created as a byproduct.</p>
New cards
13

Draw a short section of an RNA polymer (using circle, pentagon and rectangle)

A1.2.5— RNA as a polymer formed by condensation of nucleotide monomers.

Include at least three RNA nucleotides, drawn as circle (phosphate), pentagon (ribose) and rectangle (base).

Be sure the phosphate of one nucleotide is connected to the 3'C of the adjacent nucleotide.

New cards
14

Describe the structure of a DNA double helix.

A1.2.6— DNA as a double helix made of two antiparallel strands of nucleotides with two strands linked by hydrogen bonding between complementary base pairs.

Two polymers of DNA nucleotides, each with a sugar-phosphate backbone, run in antiparallel directions. Complementary DNA nitrogenous bases (A-T, C-G) form hydrogen bonds between them, binding the two polymer strands ("double") so that they wind around each other ("helix")

<p>Two polymers of DNA nucleotides, each with a sugar-phosphate backbone, run in antiparallel directions. Complementary DNA nitrogenous bases (A-T, C-G) form hydrogen bonds between them, binding the two polymer strands ("double") so that they wind around each other ("helix")</p>
New cards
15

Outline the complementary base pairing rule, including the type and number of bonds between bases.

A1.2.6— DNA as a double helix made of two antiparallel strands of nucleotides with two strands linked by hydrogen bonding between complementary base pairs.

In DNA, the nitrogenous bases of two antiparallel strands form hydrogen bonds with each other.

The complementary base pairing rule is that adenine only binds with thymine (with 2 H-bonds) and that guanine only binds with cytosine (with 3 H-bonds).

<p>In DNA, the nitrogenous bases of two antiparallel strands form hydrogen bonds with each other.</p><p>The complementary base pairing rule is that adenine only binds with thymine (with 2 H-bonds) and that guanine only binds with cytosine (with 3 H-bonds).</p>
New cards
16

Define antiparallel in relation to DNA structure.

A1.2.6— DNA as a double helix made of two antiparallel strands of nucleotides with two strands linked by hydrogen bonding between complementary base pairs.

Adjacent molecules are oriented parallel to each other but oriented in opposite directions.

In DNA, one strand runs 5' to 3' and the complementary strand runs 3' to 5'

<p>Adjacent molecules are oriented parallel to each other but oriented in opposite directions. <br><br>In DNA, one strand runs 5' to 3' and the complementary strand runs 3' to 5'</p>
New cards
17

Describe a phosphodiester bond.

A1.2.6— DNA as a double helix made of two antiparallel strands of nucleotides with two strands linked by hydrogen bonding between complementary base pairs.

A covalent bond in DNA that forms between the hydroxyl group of the 3ʹ carbon of deoxyribose and the phosphate group attached to the 5ʹ carbon of deoxyribose.

<p>A covalent bond in DNA that forms between the hydroxyl group of the 3ʹ carbon of deoxyribose and the phosphate group attached to the 5ʹ carbon of deoxyribose.</p>
New cards
18

Compare and contrast the structure of DNA and RNA.

A1.2.7— Differences between DNA and RNA.

Both are nucleic acids formed through condensation of nucleotides. Both DNA and RNA have a sugar-phosphate backbone.

RNA
ribose
single stranded
nitrogenous bases A, G, C, U
Complementary paring A-U, C-G

DNA
deoxyribose
double stranded
nitrogenous bases A, G, C, T
Complementary pairing A-T, C-G

<p>Both are nucleic acids formed through condensation of nucleotides. Both DNA and RNA have a sugar-phosphate backbone.<br><br><strong>RNA</strong><br>ribose<br>single stranded<br>nitrogenous bases A, G, C, U<br>Complementary paring A-U, C-G<br><br><strong>DNA</strong><br>deoxyribose<br>double stranded<br>nitrogenous bases A, G, C, T<br>Complementary pairing A-T, C-G</p>
New cards
19

Compare and contrast the functions of DNA and RNA.

A1.2.7— Differences between DNA and RNA.

DNA:

Passes heredity information between generations of cells

Codes for making RNA during transcription

RNA:

Codes for making proteins during translation

New cards
20

Compare and contrast the location of DNA and RNA in prokaryotic and eukaryotic cells.

A1.2.7— Differences between DNA and RNA.

Eukaryotic Cells

Both DNA and RNA are found in the nucleus.

DNA also in mitochondria and chloroplasts.

RNA also in cytoplasm and as part of ribosomes (free or bound to rough ER)

Prokaryotic Cells

Both DNA and RNA are in the cytoplasm.

DNA is clumped in a region called the nucleoid.

New cards
21

Explain the role of complementary base pairing in maintaining the DNA sequence during DNA replication.

A1.2.8— Role of complementary base pairing in allowing genetic information to be replicated and expressed.

During DNA replication, the two strands of a "parent" DNA molecule are broken apart. Each of these strands serves as a template for the creation of a new "daughter" strand. Because of the base pairing rule, the parent template strand will always code for the complementary sequence of nucleotides in the daughter strand (A to T, C to G). The complementary base paring will maintain the sequence of the DNA from generation to generation.

New cards
22

Outline the role of complementary base pairing in transmitting the genetic code in transcription.

A1.2.8— Role of complementary base pairing in allowing genetic information to be replicated and expressed.

During transcription, one of the the two strands of a DNA molecule is used as a template for the creation of an RNA strand. Because of the base pairing rule, the DNA template strand will always code for the complementary sequence of RNA nucleotides in the (A to U, C to G). The complementary base paring will maintain the sequence of the gene as mRNA is translated into protein.

New cards
23

Outline the role of complementary base pairing in transmitting the genetic code in translation.

A1.2.8— Role of complementary base pairing in allowing genetic information to be replicated and expressed.

During translation, an RNA strand is used as a template for the creation of a polypeptide. Because of the base pairing rule, the mRNA codon will only bind with the complementary tRNA anticodon (A to U, C to G). The complementary base paring ensures the correct amino acid are brought in the correct sequence to the ribosome.

New cards
24

Outline why there is a limitless diversity of DNA base sequences.

A1.2.9— Diversity of possible DNA base sequences and the limitless capacity of DNA for storing Information.

There are four nitrogenous bases in DNA (A, T, C and G). These 4 bases are components of nucleotides that can form a DNA molecule in any order and of any length.

New cards
25

Define universal in relation to the genetic code.

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

Universal means that the characteristic is shared by all life.

A universal generic code means that all life uses essentially the same code when translating information stored in genes into a polypeptide.

New cards
26

Outline why conservation of the genetic code across all forms of life is evidence of common ancestry.

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

Using inductive reasoning, it can be concluded that the use of the same genetic code across all forms of life indicates that all organisms inherited the use of the code from a common ancestor. The alternative, that all forms of life independently developed use of the same genetic code, is an illogical hypothesis.

New cards
27

Identify and label the 5’ and 3’ ends on a DNA or RNA diagram.

AHL A1.2.11— Directionality of RNA and DNA.

Each end of DNA nucleotide has a number. One end is referred to as 5' (five prime) and the other end is referred to as 3' (three prime). The 5' and 3' designations refer to the number of the carbon atom in a deoxyribose sugar molecule. The 5' end is identified by the presence of the phosphate group and the 3' end is identified as ending in the pentose sugar (ribose or deoxyribose).

In DNA, one strand will run from 5' to 3' and the complementary strand will run anti-parallel, from 3' to 5'.

<p>Each end of DNA nucleotide has a number. One end is referred to as 5' (five prime) and the other end is referred to as 3' (three prime). The 5' and 3' designations refer to the number of the carbon atom in a deoxyribose sugar molecule. The 5' end is identified by the presence of the phosphate group and the 3' end is identified as ending in the pentose sugar (ribose or deoxyribose). <br><br>In DNA, one strand will run from 5' to 3' and the complementary strand will run anti-parallel, from 3' to 5'.</p>
New cards
28

Compare and contrast the structures of purines and pyrimidines.

AHL A1.2.12— Purine-to-pyrimidine bonding as a component of DNA helix stability.

Both purines and pyrimidines are nitrogenous bases of DNA and RNA.

Pyrimidine: single ring nitrogenous bases
Cytosine
Thymine
Uracil

Purine: double ring nitrogenous bases
Guanine
Adenine

<p>Both purines and pyrimidines are nitrogenous bases of DNA and RNA. <br><br><strong>Pyrimidine:</strong> single ring nitrogenous bases<br>Cytosine<br>Thymine<br>Uracil <br><br><strong>Purine:</strong> double ring nitrogenous bases<br>Guanine <br>Adenine</p>
New cards
29

State that in DNA, a purine forms hydrogen bonds with a pyrimidine.

AHL A1.2.12— Purine-to-pyrimidine bonding as a component of DNA helix stability.

In DNA, a purine complementary base pairs to a pyrimidine using hydrogen bonds

In DNA and RNA, guanine bonds with cytosine with three hydrogen bonds.

In DNA, adenine bonds with thymine with two hydrogen bonds.
In RNA, adenine bonds with uracil with two hydrogen bonds

New cards
30

Given a diagram of DNA, identify the four bases of DNA based on purine or pyrimidine and the number of hydrogen bonds.

AHL A1.2.12— Purine-to-pyrimidine bonding as a component of DNA helix stability.

Purines have two rings. If it can form 2 H-bonds it is adenine and if it can form 3 H-bonds it is guanine.

Pyrimidines have one ring. If it can form 2 H-bonds it is thymine and if it can form 3 H-bonds it is cytosine.

New cards
31

Describe the structure of eukaryotic DNA and associated histone proteins during interphase (chromatin).

AHL A1.2.13— Structure of a nucleosome.

To compact DNA while regulating gene accessibility for transcription, eukaryotic organisms organize their genomes:

1. DNA double helix.
2. DNA wraps around histone proteins, forming nucleosomes and the "beads on a string" structure.
3. Multiple nucleosomes wrap into a fibre (chromatin).
4. Supercoiling of the chromatin produces the chromosome (during mitosis and meiosis). Supercoiling refers to the repeated twisting and winding of the DNA strand. Supercoiling functions to reduce the space required for DNA packaging, allowing for more compact storage of DNA.

<p>To compact DNA while regulating gene accessibility for transcription, eukaryotic organisms organize their genomes:<br><br>1. DNA double helix.<br>2. DNA wraps around histone proteins, forming nucleosomes and the "beads on a string" structure.<br>3. Multiple nucleosomes wrap into a fibre (chromatin).<br>4. Supercoiling of the chromatin produces the chromosome (during mitosis and meiosis). Supercoiling refers to the repeated twisting and winding of the DNA strand. Supercoiling functions to reduce the space required for DNA packaging, allowing for more compact storage of DNA.</p>
New cards
32

Draw and label the structure of a nucleosome, including the H1 protein, the octamer core proteins, linker DNA and two wraps of DNA.

AHL A1.2.13— Structure of a nucleosome.

The nucleosome is the basic unit of DNA packaging in eukaryotes. Each nucleosome is composed of two turns of DNA wrapped around a group of eight histone proteins (called an octamer core). Each nucleosome connects to the adjacent nucleosomes through another type of histone protein (called the H1) and a region of "linker" DNA.

<p>The nucleosome is the basic unit of DNA packaging in eukaryotes. Each nucleosome is composed of two turns of DNA wrapped around a group of eight histone proteins (called an octamer core). Each nucleosome connects to the adjacent nucleosomes through another type of histone protein (called the H1) and a region of "linker" DNA.</p>
New cards
33

Define a histone.

AHL A1.2.13— Structure of a nucleosome.

Histone: a protein molecule around which DNA is tightly coiled. There are two of each kind of histone in a nucleosome.

New cards
34

Define a nucleosome.

AHL A1.2.13— Structure of a nucleosome.

Nucleosome: Linker DNA coiled twice around eight histone proteins. DNA is attracted to the histones because DNA is negatively charged and the histones are positively charged

<p>Nucleosome: Linker DNA coiled twice around eight histone proteins. DNA is attracted to the histones because DNA is negatively charged and the histones are positively charged</p>
New cards
35

Distinguish between exons and introns

AHL A1.2.13— Structure of a nucleosome.

Exons: fragments that code for proteins

Introns: non-coding fragments

New cards
36

Define a pseudogene.

AHL A1.2.13— Structure of a nucleosome.

Pseudogene: sections of DNA that don’t appear to have a coding function

New cards
37

Define a polymorphism.

AHL A1.2.13— Structure of a nucleosome.

Polyphormism: specific regions of DNA that show significant variation between people

New cards
38

Outline the significance of DNA packaging.

AHL A1.2.13— Structure of a nucleosome.

DNA packaging is essential because the nucleus is microscopic, so DNA is paired with a type of protein called a histone.

DNA is inaccessible to transcription enzymes when wrapped around histones & in even more elaborate structures. This regulates the transcription process, allowing only certain areas of the DNA molecule to be involved in protein synthesis.

<p>DNA packaging is essential because the nucleus is microscopic, so DNA is paired with a type of protein called a histone.</p><p>DNA is inaccessible to transcription enzymes when wrapped around histones &amp;  in even more elaborate structures. This regulates the transcription process, allowing only certain areas of the DNA molecule to be involved in protein synthesis.</p>
New cards
39

Define DNA profiling.

AHL A1.2.13— Structure of a nucleosome.

DNA Profiling: A procedure that analyzes DNA fragments to determine whether they come from a specific individual

New cards
40

Define a loci.

AHL A1.2.13— Structure of a nucleosome.

Loci: Location of a gene on a chromosome

New cards
41

List the types of DNA sequences.

AHL A1.2.13— Structure of a nucleosome.

Highly repetitive Sequences (45%):  Groups of repetitive DNA that do not have coding functions and are transposable (can move from one genome location to another)

Protein-coding Genes (2%): Single-copy genes that have coding functions

Structural DNA (20%): Highly coiled DNA that does not have a coding function; occurs around the centromere & near the ends of chromosomes at the telomeres

Short Tandem Repeats: a group of 13 very specific loci that are short, repeating sequences of DNA (normally 2-5 base pairs)

<p><strong>Highly repetitive Sequences (45%): </strong>&nbsp;Groups of repetitive DNA that do not have coding functions and are transposable (can move from one genome location to another)</p><p><strong>Protein-coding Genes (2%):</strong> Single-copy genes that have coding functions</p><p><strong>Structural DNA (20%):</strong> Highly coiled DNA that does not have a coding function; occurs around the centromere &amp; near the ends of chromosomes at the telomeres</p><p><strong>Short Tandem Repeats:</strong> a group of 13 very specific loci that are short, repeating sequences of DNA (normally 2-5 base pairs)</p>
New cards
42

State the experimental question being tested in the Hershey and Chase experiment.

AHL A1.2.14— Evidence from the Hershey–Chase experiment for DNA as the genetic material.

Both protein and nucleic acids are part of the eukaryotic cell nucleus. Prior to 1952 it was unknown which molecule type was responsible for passing genetic information between generations.

Hershey and Chase designed an experiment to test whether proteins or DNA is the hereditary material.

New cards
43

Outline the procedure of the Hershey and Chase experiment.

AHL A1.2.14— Evidence from the Hershey–Chase experiment for DNA as the genetic material.

Hershey and Chase (H&C) performed their experiments on viruses that infect bacteria called bacteriophages.

They incorporated radioactive isotopes of phosphorus and sulfur into phages.

DNA contains phosphorus, but not sulfur, whereas protein contains sulfur, but not phosphorus. Therefore, when H &C marked phages with radioactive isotopes of those elements, they placed separate, distinguishable tags on the protein and DNA parts of the phages.

They allowed the phages to replicate by infecting bacteria.

By tracking the location of the radioactive tags, H&C showed that phages only injected their DNA into host bacteria, and that the DNA served as the replicating genetic element of phages. The phages did not inject their protein coats into the bacteria; the protein coats remained outside the bacteria adhered to the bacterial membranes.

<p>Hershey  and Chase (H&amp;C) performed their experiments on viruses that infect  bacteria called bacteriophages.</p><p>They incorporated radioactive isotopes of phosphorus and sulfur into phages.</p><p>DNA contains phosphorus, but not  sulfur, whereas protein contains sulfur, but not phosphorus. Therefore,  when H &amp;C marked phages with radioactive isotopes of those elements,  they placed separate, distinguishable tags on the protein and DNA parts  of the phages.</p><p>They allowed the phages to replicate by infecting  bacteria.</p><p>By tracking the location of the radioactive tags, H&amp;C  showed that phages only injected their DNA into host bacteria, and that  the DNA served as the replicating genetic element of phages. The phages did not inject their protein coats into the bacteria; the protein coats  remained outside the bacteria adhered to the bacterial membranes.</p>
New cards
44

Explain how the results of the Hershey and Chase experiment supported the notion of nucleic acids as the genetic material.

AHL A1.2.14— Evidence from the Hershey–Chase experiment for DNA as the genetic material.

The molecule of heredity must be carried from generation to generation.

When Hershey and Chase demonstrated that radioactively tagged DNA is present across generations of bacteriophages and that radioactively tagged protein is not, they demonstrated that nucleic acids must be the genetic material.

<p>The molecule of heredity must be carried from generation to generation.</p><p>When  Hershey and Chase demonstrated that radioactively tagged DNA is present  across generations of bacteriophages and that radioactively tagged protein is not, they demonstrated that nucleic acids must be the genetic  material.</p>
New cards
45

Outline the use of radioisotopes as research tools.

AHL A1.2.14— Evidence from the Hershey–Chase experiment for DNA as the genetic material.

Hershey and Chase relied on radioactive isotopes of phosphorus and sulfur in their experiment to determine that DNA is the genetic material.

Because they give off energy as they convert to a more stable form, the isotopes allow biologists to track molecules as they move through biological systems.

New cards
46

Define virus and bacteriophage

AHL A1.2.14— Evidence from the Hershey–Chase experiment for DNA as the genetic material.

Virus: DNA or RNA enclosed by a protective protein coat that infects cells

Bacteriophage: viruses that only infect bacteria cells

<p>Virus:  DNA or RNA enclosed by a protective protein coat that infects cells</p><p>Bacteriophage:  viruses that only infect bacteria cells</p>
New cards
47

Describe the implications of Chargaff’s data that showed a 1:1 ratio of purine to pyrimidine in a sample of DNA.

AHL A1.2.15— Chargaff’s data on the relative amounts of pyrimidine and purine bases across diverse life forms.

Chargaff's research revealed the percentage of each base (A, T, C, G) found in different species DNA was different. He determined that there are equal numbers of A and T bases and G and C bases in a DNA sample, resulting in a 1:1 ratio of purines to pyrimidines in DNA.

Chargaff did not use inductive reasoning to then infer the pattern that A binds to T and C binds to G. The complementary base pairing rule was inferred by Watson and Crick.

New cards
48

Explain the role of falsifiability in determining the structure and function of DNA.

AHL A1.2.15— Chargaff’s data on the relative amounts of pyrimidine and purine bases across diverse life forms.

To falsify means that a statement, theory or hypothesis is proven to be wrong.

Because he found that the amount of one base wasn't equal to the amount of the three other bases , Chargaff's data falsified the earlier hypothesis that DNA had a tetranucleotide structure. In this proposed model of DNA, the four bases occur in equal amounts in four strands arranged with the bases facing outward.

<p>To falsify means that a statement, theory or hypothesis is proven to be wrong.</p><p>Because he found that the amount of one base wasn't equal to the amount of the three other bases , Chargaff's data falsified the earlier hypothesis that DNA had a tetranucleotide structure. In this proposed model of DNA, the four bases occur in equal amounts in four strands arranged with the bases facing outward.</p>
New cards

Explore top notes

note Note
studied byStudied by 150 people
Updated ... ago
5.0 Stars(4)
note Note
studied byStudied by 48 people
Updated ... ago
5.0 Stars(2)
note Note
studied byStudied by 12 people
Updated ... ago
5.0 Stars(1)
note Note
studied byStudied by 2 people
Updated ... ago
5.0 Stars(1)
note Note
studied byStudied by 20 people
Updated ... ago
5.0 Stars(1)
note Note
studied byStudied by 6 people
Updated ... ago
5.0 Stars(1)
note Note
studied byStudied by 4 people
Updated ... ago
5.0 Stars(1)
note Note
studied byStudied by 41 people
Updated ... ago
5.0 Stars(4)

Explore top flashcards

flashcards Flashcard28 terms
studied byStudied by 42 people
Updated ... ago
5.0 Stars(1)
flashcards Flashcard29 terms
studied byStudied by 1 person
Updated ... ago
5.0 Stars(1)
flashcards Flashcard46 terms
studied byStudied by 2 people
Updated ... ago
5.0 Stars(1)
flashcards Flashcard28 terms
studied byStudied by 34 people
Updated ... ago
5.0 Stars(3)
flashcards Flashcard27 terms
studied byStudied by 16 people
Updated ... ago
5.0 Stars(1)
flashcards Flashcard33 terms
studied byStudied by 17 people
Updated ... ago
4.5 Stars(2)
flashcards Flashcard62 terms
studied byStudied by 7 people
Updated ... ago
5.0 Stars(1)
flashcards Flashcard324 terms
studied byStudied by 7 people
Updated ... ago
5.0 Stars(1)