14. A1.2 Nucleic Acids

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Last updated 3:53 PM on 9/4/26
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33 Terms

1
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State the 2 primary functions of nucleic acids

  1. Hereditary material by replicating and transferring genetic information to the next generation

  2. Code for protein production in the process of gene expression


2
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State the two types of nucleic acids used in cells

DNA - deoxyribonucleic acid

  • Passes heredity information between generations of cells

  • Codes for making RNA during transcription

RNA - ribonucleic acid

  • Codes for making proteins during translation

  • mRNA, rRNA, and tRNA are the three main types of RNA involved in protein synthesis


3
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Outline the meaning and implication of DNA being the genetic material of all living organisms

Evidence of universal common ancestry of life

4
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State why RNA viruses do not falsify the claim that all living things use DNA as the genetic material

Some viruses use RNA as their genetic material.

However, because viruses are not made of cells, they are not considered to be living.

5
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​List the 3 components of a nucleotide. 

  • nucleotides (monomer), nucleic acid (polymer)

  • Nucleotides are composed of:

    • nitrogenous base

      • Adenine (A)

      • Thymine (T)

      • Cytosine (C)

      • Guanine (G)

      • Uracil (U)

    • a five-carbon "pentose" sugar (ribose or deoxyribose)

    • A negatively charged phosphate group


6
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Draw the basic structure of a single nucleotide (using circle, pentagon and rectangle), labelling the carbons of the pentose sugar.

  • nitrogenous base connecting to carbon-1

  • carbon-5 branches out of the ring

  • phosphate group connecting to carbon-5


7
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Define “backbone” as related to nucleic acid structure

  • Phosphodiester bond:

    • Each phosphate group is covalently bonded to the 3rd carbon of the pentose sugar before it and to the 5th carbon of the next sugar


  • Nucleotides connect to form a strong sugar-phosphate backbone, from 5’ to 3’

  • 5’ terminal = the end where the last phosphate is

  • 3’ terminal = the end where a phosphate could link to the 3rd carbon

DNA: double helix, 2 sugar phosphate backbones that run antiparallel

RNA: one sugar-phosphate backbone


Function: provides structural support

The sharing of electrons in the covalent bond between sugar and phosphate provides strength to the structure.

The strength maintains the nucleotides in their specific sequence, (for storing, replication and expression of genetic information)

<ul><li><p>Phosphodiester bond:</p><ul><li><p><span style="background-color: transparent;">Each phosphate group is <u>covalently bonded </u>to the 3<sup>rd</sup> carbon of the pentose sugar before it and to the 5<sup>th</sup> carbon of the next sugar</span></p></li></ul></li></ul><p></p><ul><li><p><span style="background-color: transparent;">Nucleotides connect to form a <strong><u>strong sugar-phosphate backbone,</u> from 5’ to 3’</strong></span></p></li><li><p><span style="background-color: transparent;">5’ terminal = the end where the last phosphate is </span></p></li><li><p><span style="background-color: transparent;">3’ terminal = the end where a phosphate could link to the 3<sup>rd</sup> carbon</span></p></li></ul><p><strong>DNA:</strong> double helix, 2 sugar phosphate backbones that run antiparallel</p><p><strong>RNA: </strong>one sugar-phosphate backbone </p><p></p><p><span style="background-color: transparent;">Function: provides <strong>structural support</strong></span></p><p><span style="background-color: transparent;">The sharing of electrons in the covalent bond between sugar and phosphate provides strength to the structure.</span></p><p><span style="background-color: transparent;">The strength maintains the nucleotides in their specific sequence, (for storing, replication and expression of genetic information)  </span></p>
8
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State a similarity and a difference between the nitrogenous bases


9
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Outline how the sequence of bases in a nucleic acid serves as a ‘code’.

CODON = a set of three bases that codes for an amino acid


10
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Define gene

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

11
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Describe the structure of a DNA double helix

  • The two strands of DNA is antiparallel
    = they are parallel but run in opposite directions

    • At each end of the double helix, one strand is 5’ and the other is 3’


  • The two strands ‘twist’ to form a double-helix structure

  • The strands are held together by hydrogen bond between the complementary base pairings.

    • Adenine – Thymine ( 2 H-bonds)

    • Guanine – Cytosine (3 H-bonds)


12
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Compare and contrast the structures of DNA and RNA


13
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Compare and contrast the functions of DNA and RNA


14
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Compare and contrast the location of DNA and RNA in prokaryotic and eukaryotic cells


15
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Outline the role of complementary base pairing in maintaining the DNA sequence during DNA replication

DNA replication occurs before cell division.

DNA replication is "semi-conservative"

  • each of the original DNA strands serves as a template for the creation of a new strand.

  • DNA polymerase III builds the new strand by "reading" the template and adding the complementary DNA nucleotide.

  • This results in the newly built strand having the same sequence of bases as the other template strand.

In this way, replication builds two identical DNA molecules, each with one original and one new strand.


16
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Understanding translation and transciption in DNA


17
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Outline the role of complementary base pairing in transmitting the genetic code in transcription

Transcription is the synthesis of RNA using a DNA template.

The enzyme RNA polymerase builds an RNA strand by "reading" the DNA template and adding the complementary RNA nucleotide.

  • RNA strand having the complementary sequence of bases as the DNA, thereby maintaining the information stored in the sequence of nucleotides of the code.


18
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Outline the role of complementary base pairing in transmitting the genetic code in translation

Translation is the synthesis of a polypeptide from mRNA.

  • ribosome builds a polypeptide by reading the mRNA template and binding the coded amino acid to the polypeptide chain

  • amino acids are brought to the ribosome by a tRNA.

  • tRNA forms a temporary bond to the mRNA using complementary base pairing



19
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Outline why there is a limitless diversity of DNA base sequences

  • 4 bases = This can code for 20 different amino acids

  • Polypeptide length can range from 20-2000 amino acids

  • Average length of a polypeptide is around 300 = 20300 = ~ 1.4… x 10390 possible polypeptide or ~1.24 x 10723 different base sequence

  • The arrangement of any base sequence is possible and a gene can also be of any length

  • And can code for a limitless possibility of polypeptide with difference amino acid sequences which leads to a very diverse range of genes

  • There is an almost limitless capacity of DNA for storing a wide range of data with great economy as it only needs 4 different bases, can be store within such a small structure of a cell and can be easily replicated


20
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Outline why conservation of the genetic code across all forms of life is evidence of common ancestry

  • All organisms using the same genetic material – DNA (excluding virus)

  • All organisms uses the complementary base pairing (prokaryote uses uracil in their DNA instead of thymine)

  • And all organisms replicate their DNA, transcribed and translate their DNA into protein

  • Most using the same 3-base codon to code for amino acid sequence

  • Suggesting a universal common ancestry (despite the hypothesis that RNA is the first genetic material to evolve)

  • *In very limited organisms, some codons have different meaning, which raised some hypothesis that genetic code might not be universal, or that codons have occurred after the origin of life


Universal definition in this context: The genetic code is universal, which means that all living organisms use this same genetic code. Even viruses use this same code


21
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Directionality of RNA & DNA

  • DNA strands are antiparallel and DNA & RNA strands are produced from the 5’ to 3’ direction

  • Meaning that DNA nucleotides & RNA nucleotides can only be added to the 3’ end of a growing strand 

  • This is because the condensation reaction to form the sugar phosphate bone occurs between the hydroxyl group of phosphate (attached to the 5’ carbon) in the new nucleotide and the hydroxyl group attached to the 3’ carbon

  • The addition of the nucleotide during RNA/DNA synthesis involve the cleavage of two phosphate group from the dNTP (deoxyribose nucleoside triphosphate) which releases energy to form the phosphodiester bond


22
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Identify and label the 5’ and 3’ ends of the daughter DNA strands on a diagram of the DNA replication fork

Since both strands of DNA are used as a template during DNA replication, there are differences in how replication occurs on each strand

  • Leading strand: the new DNA polymer
    3' end is at the opening of the replication fork

  • Lagging strand: the new DNA polymer
    3' end is opposite the opening of the replication fork


<p>Since both strands of DNA are used as a template during DNA replication, there are differences in how replication occurs on each strand</p><ul><li><p>Leading strand: the new DNA polymer<br>3' end is at the opening of the replication fork</p></li><li><p>Lagging strand: the new DNA polymer<br>3' end is opposite the opening of the replication fork</p><img src="https://assets.knowt.com/user-attachments/14ca84fc-d5fc-41eb-94d5-f45cf08d269f.png" data-width="50%" data-align="center" style="display: block; width: 50%; margin-left: auto; margin-right: auto;"><div data-youtube-video=""><iframe width="640" height="480" allowfullscreen="true" autoplay="false" disablekbcontrols="false" enableiframeapi="false" endtime="0" ivloadpolicy="0" loop="false" modestbranding="false" origin="" playlist="" rel="1" src="https://www.youtube.com/embed/Qqe4thU-os8?rel=1" start="0"></iframe></div></li></ul><p></p>
23
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Identify and label the 5’ and 3’ ends of RNA on a diagram of the transcription bubble


24
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Outline the impact of DNA directionality on translation


25
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Compare and contrast the structures of purines and pyrimidines

Compare: All have nitrogen

Contrast: Purines = double ring structures, Pyrimidines = single ring structures

26
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Purine to pyrimidine bonding

  • A can only pair with T & C can only pair with G

  • The bases pair upside down with each other

  • The complementary base pairs have
    equal length between the two strands


27
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State two consequences of purine-to-pyrimidine bonding on the structure of DNA


  1. When a pyrimidine is paired with a purine, the width dimension of both pairs is identical. This means that the DNA sugar-phosphate backbones have a consistent diameter throughout the entire molecule.

  2. The hydrogen bonding between the double helix helps to greatly stabilise the structure.


28
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Function of nucleosomes

  • Eukaryotic Cells: have histones to help package DNA into nucleosomes

  • The nucleosome contributes to supercoiling of DNA and is an adaptation that allows the packing of large genome within a small space in eukaryotes

  • It also protects the DNA from damage and allows chromosomes to move during mitosis & meiosis without entanglement.


29
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Structure of nucleosomes

A nucleosome is made up of

  • eight histone proteins (octamer) with DNA coiled around it

  • A linker DNA connects one nucleosome to the other

  • Another histone protein H1, binds the DNA to the core particle 

  • Forming a beads-on-a-string structure (chromatosomes) – genes can still be activated


<p><span style="background-color: transparent;">A nucleosome is made up of</span></p><ul><li><p><span style="background-color: transparent;">eight histone proteins (octamer) with DNA coiled around it</span></p></li></ul><ul><li><p><span style="background-color: transparent;">A linker DNA connects one nucleosome to the other</span></p></li><li><p><span style="background-color: transparent;">Another histone protein H1, binds the DNA to the core particle&nbsp;</span></p></li><li><p><span style="background-color: transparent;">Forming a beads-on-a-string structure (chromatosomes) – genes can still be activated </span></p></li></ul><p></p>
30
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Draw and label the structure of a nucleosome, including the H1 protein, the octamer core proteins, linker DNA and two wraps of DNA


31
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DNA packaging

DNA packing with nucleosomes occurs only in eukaryotic cells.

Prokaryotic cell DNA remains "naked" meaning it does not wrap around histone proteins.

  • The H1 proteins will interact and form a ring of 6 nucleosomes which will coil and stack upon one another together to form a fibre which is 30nm in diameter (also called a solenoid)  - like a hollow tube

  • This 30nm fibre form further loops that coils around each other and condense into chromatins (700nm) – part of the chromosome

  • During prophase of mitosis and meiosis, DNA condenses even more by supercoiling into chromosomes (D2.1.6). Condensing DNA into chromosomes prevents DNA tangling and damage during cell division.



<p>DNA packing with nucleosomes occurs only in eukaryotic cells.</p><p class="p1">Prokaryotic cell DNA remains "naked" meaning it does not wrap around histone proteins.</p><img src="https://assets.knowt.com/user-attachments/9dd75f8b-b657-4c30-a1a6-09f91f56a527.png" data-width="50%" data-align="center" style="display: block; width: 50%; margin-left: auto; margin-right: auto;"><ul><li><p><span style="background-color: transparent;">The H1 proteins will interact and form a <strong>ring of 6 nucleosomes</strong> which will coil and stack upon one another together to form a <strong>fibre which is 30nm in diameter</strong> (also called a <strong>solenoid</strong>)&nbsp; - like a hollow tube</span></p></li><li><p><span style="background-color: transparent;">This 30nm fibre form further loops that coils around each other and <strong>condense into chromatins</strong> (700nm) – part of the chromosome</span></p></li><li><p>During prophase of mitosis and meiosis, DNA condenses even more by supercoiling into <strong>chromosomes</strong> (D2.1.6). Condensing DNA into chromosomes prevents DNA tangling and damage during cell division.</p></li></ul><p></p><img src="https://assets.knowt.com/user-attachments/e828b444-e25b-49e6-8d8e-95afa46d463d.png" data-width="50%" data-align="center" style="display: block; width: 50%; margin-left: auto; margin-right: auto;"><p></p>
32
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Outline the procedure of the Hershey and Chase experiment

  • 1940s – scientists were unsure whether protein or DNA was the hereditary material

  • Proteins were more favoured as it can produce a great variety due to 20 naturally occurring subunits instead of only 4 nucleotide subunits

  • Experiment

  • It was known then that virus:

    • Binds to host cell and inject their genetic material into cells

    • Its non-genetic material remain outside of cell

    • The infected cell will then produce large numbers of virus until it bursts and releases the viruses into the environment

    • So the substance injected inside of the virus must contain the hereditary material for producing new viruses

    • Virus has a protein coat, and a DNA core

  1. They choose to investigate T2 bacteriophage that is a very simple virus that consists only of a protein coat (capsid) and a strand of DNA

  2. Grew one batch of virus with radioactive sulfur (35S) --> label proteins since protein only have sulfur and DNA does not have.

  3. Grow another batch of virus with radioactive phosphorus (32P) —> label the DNA but not the protein

  4. Test: if the radioactive material is inside the bacteria or was left outside.

  5. They then infect the bacteria with radioactively-labelled virus (separately)

  6. Separate the bacteria and virus through centrifugation as the

  7. Heavier bacteria sink to the bottom to form a pellet

  8. The rest of the virus remains in supernatant (top layer liquid)


  • If protein was the genetic material, then they should find more 35S protein in the bacteria (pellet) & no radioactive in the supernatant

  • If DNA is the genetic material, then they should find more 35P DNA in the bacteria (pellet)  & no radioactive protein in the supernatant.


  • Results:

  • 35S remain in supernatant

  • 32P in pellet – meaning that DNA is in the bacteria


33
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Describe implications of Chargaff’s data that showed a 1:1 ratio of purine to pyrimidine in a sample of DNA

  • Levene was a scientist who discovered nucleotides and proposed the structure of nucleic acid – Tetranucleotide Hypothesis

  • Another scientist – Chargaff decided to calculate the different amount of purine and pyrimidines in different organisms

  • He found that the 4 bases was not equal in amount as proposed by the tetranucleotide hypothesis

  • Instead, he found that the amount of 

    • adenine = thymine & cytosine = guanine 

  • Although Chargaff did not figure out the structure of DNA, 

  • Chargaff’s data falsified the tetranucleotide hypothesis that there was a repeating sequence of the four bases of DNA