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Seven Quest Topic Five: Nucleic Acids Commack High School HL Biology

Topic Five: Nucleic Acids student notes

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Page 1: Topic Five: Nucleic Acids student notes

Com

mack

High

School HL

Biology

Topic

Five: Nucl

eic Acid

s

Seven Quest

Page 2: Topic Five: Nucleic Acids student notes

Topic Five Nucleic Acids Name:

Understandings, Applications and Skills (This is what you maybe assessed on)

Statement Guidance2.7.U1 The replication of DNA is semi-conservative and depends

on complementary base pairing.

2.7.U2 Helicase unwinds the double helix and separates the two strands by breaking hydrogen bonds.

2.7.U3 DNA polymerase links nucleotides together to form a new strand, using the pre-existing strand as a template.

The different types of DNA polymerase do not need to be distinguished.

2.7.U4 Transcription is the synthesis of mRNA copied from the DNA base sequences by RNA polymerase.

2.7.U5 Translation is the synthesis of polypeptides on ribosomes.

2.7.U6 The amino acid sequence of polypeptides is determined by mRNA according to the genetic code.

2.7.U7 Codons of three bases on mRNA correspond to one amino acid in a polypeptide.

2.7.U8 Translation depends on complementary base pairing between codons on mRNA and anticodons on tRNA.

2.7.A1 Use of Taq DNA polymerase to produce multiple copies of DNA rapidly by the polymerase chain reaction (PCR).

2.7.A2 Production of human insulin in bacteria as an example of the universality of the genetic code allowing gene transfer between species.

2.7.S1 Use a table of the genetic code to deduce which codon(s) corresponds to which amino acid.

2.7.S2 Analysis of Meselson and Stahl’s results to obtain support for the theory of semi-conservative replication of DNA.

2.7.S3 Use a table of mRNA codons and their corresponding amino acids to deduce the sequence of amino acids coded by a short mRNA strand of known base sequence.

2.7.S4 Deducing the DNA base sequence for the mRNA strand.

Understandings, Applications and Skills (This is what you may be assessed on)

Page 3: Topic Five: Nucleic Acids student notes

Topic Five Nucleic Acids Name:

Statement Guidance7.1 U.1 Nucleosomes help to supercoil the DNA.

7.1 U.2 DNA structure suggested a mechanism for DNA replication.

7.1 U.3 DNA polymerases can only add nucleotides to the 3’ end of a primer.

7.1 U.4 DNA replication is continuous on the leading strand and discontinuous on the lagging strand.

Details of DNA replication differ between prokaryotes and eukaryotes. Only the prokaryotic system is expected.

7.1 U.5 DNA replication is carried out by a complex system of enzymes.

The proteins and enzymes involved in DNA replication should include helicase, DNA gyrase, single strand binding proteins, DNA primase and DNA polymerases I and III.

7.1 U.6 Some regions of DNA do not code for proteins but have other important functions.

The regions of DNA that do not code for proteins should be limited to regulators of gene expression, introns, telomeres and genes for tRNAs.

7.1 A.1 Rosalind Franklin’s and Maurice Wilkins’ investigation of DNA structure by X-ray diffraction.

7.1 A.2 Use of nucleotides containing deoxyribonucleic acid to stop DNA replication in preparation of samples for base sequencing.

7.1 A.3 Tandem repeats are used in DNA profiling.

7.1 S.1 Analysis of results of the Hershey and Chase experiment providing evidence that DNA is the genetic material.

7.1 S.2 Utilization of molecular visualization software to analyse the association between protein and DNA within a nucleosome.

Understandings, Applications and Skills (This is what you may be assessed on)

Statement Guidance7.2 U.1 Transcription occurs in a 5’ to 3’ direction. RNA polymerase adds the 5´ end of the

free RNA nucleotide to the 3´ end of the growing mRNA molecule.

7.2 U.2 Nucleosomes help to regulate transcription in eukaryotes.

7.2 U.3 Eukaryotic cells modify mRNA after transcription.

7.2 U.4 Splicing of mRNA increases the number of different proteins an organism can produce.

Page 4: Topic Five: Nucleic Acids student notes

Topic Five Nucleic Acids Name:

7.2 U.5 Gene expression is regulated by proteins that bind to specific base sequences in DNA.

7.2 U.6 The environment of a cell and of an organism has an impact on gene expression.

7.2 A.1 The promoter as an example of non-coding DNA with a function.

7.2 S.1 Analysis of changes in the DNA methylation patterns.

Understandings, Applications and Skills (This is what you maybe assessed on)

Statement Guidance7.3 U.1 Initiation of translation involves assembly of the

components that carry out the process.Examples of start codons are not required. Names of the tRNA binding sites are expected as well as their roles.

7.3 U.2 Synthesis of the polypeptide involves a repeated cycle of events.

7.3 U.3 Disassembly of the components follows termination of translation.

Examples of stop codons are not required.

7.3 U.4 Free ribosomes synthesize proteins for use primarily within the cell.

7.3 U.5 Bound ribosomes synthesize proteins primarily for secretion or for use in lysosomes.

7.3 U.6 Translation can occur immediately after transcription in prokaryotes due to the absence of a nuclear membrane.

7.3 U.7 The sequence and number of amino acids in the polypeptide is the primary structure.

7.3 U.8 The secondary structure is the formation of alpha helices and beta pleated sheets stabilized by hydrogen bonding.

7.3 U.9 The tertiary structure is the further folding of the polypeptide stabilized by interactions between R groups.

Polar and non-polar amino acids are relevant to the bonds formed between R groups.

7.3 U.10 The quaternary structure exists in proteins with more than one polypeptide chain.

Quaternary structure may involve the binding of a prosthetic group to form a conjugated protein.

7.3 A.1 tRNA-activating enzymes illustrate enzyme–substrate specificity and the role of phosphorylation.

7.3 S.1 Identification of polysomes in electron micrographs of prokaryotes and eukaryotes.

7.3 S.2 The use of molecular visualization software to analyse the structure of eukaryotic ribosomes and a tRNA molecule.

Page 5: Topic Five: Nucleic Acids student notes

Topic Five Nucleic Acids Name:

7.1 S.1 Analysis of results of the Hershey and Chase experiment providing evidence that DNA is the genetic material. (Slide 8)

Use the hyperlink https://www.youtube.com/watch?v=Wxvd55XGrqc or the ppt. to view an animation about the Hersey-Chase experiment then answer the 3 questions below:

1. Explain why sulphur was used in one experiment and phosphorus in the other.

2. Explain why most of the radioactive sulphur (35S) was found in the supernatant.

3. Explain why little of the radioactive phosphorous (32P) was found in the supernatant, i.e. most remained in the pellet.

7.1 A.1 Rosalind Franklin’s and Maurice Wilkins’ investigation of DNA structure by X-ray diffraction.

When X-rays are directed at a material some is scattered by the material. This scattering is known as diffraction. For X-ray diffraction to work well the material ideally should be crystallised so that the repeating pattern causes diffraction to occur in a regular way. DNA cannot be crystallised but the molecules were arranged regularly enough for the technique to work. (pg. 88)

4. Use the animation (http://www.dnalc.org/view/15874-Franklin-s-X-ray.html) to understand how to interpret the Rosalind Franklin’s and Maurice Wilkins’ X-ray diffraction photographs of DNA. To the right is an example of a X-ray diffraction photograph of DNA.

n.b. Your answers to the questions below may well be helped by diagrams.

a. What can be deduced from the X-shaped pattern?

Nature of Science: Rosalind Franklin’s careful observation and interpretation of the photographic evidence was crucial to Crick’s and Watson’s successful discovery of the structure of DNA. Her work and her calculations were shown to Crick and Watson without her permission and they subsequently published their model before she had an opportunity to publish her work. Her work is now is widely recognised as being as important to the discovery of DNA as Crick and Watson, but unfortunately she has never shared in the Nobel prize awarded to Crick and Watson as Nobel prizes cannot be given posthumously (Franklin died in 1958 aged just 37).

7.1 U.2 DNA structure suggested a mechanism for DNA replication.

5. What are the side of the ladder that makes up DNA (Slide 10)

http://undsci.berkeley.edu/images/us101/xray.jpg

Page 6: Topic Five: Nucleic Acids student notes

Topic Five Nucleic Acids Name:

6. Outline the basic components of a nucleotide with a drawing (Slides 11 &12)

7. Describe how the complementary base pairing in DNA comes together below. (slide 14) Explain why it is only possible for cytosine to pair with guanine and adenine to pair with thymine in terms of the sugar deoxyribose in DNA (Slides 17-18)

7.1 U.1 Nucleosomes help to supercoil the DNA.

8. Explain why a Prokaryotic DNA is described as being ‘naked’ (Slide 19).

9. In the space below, draw and label the structure of a simplified nucleosome, including the H1 linker and octamer (which consists of two copies of four different types of histone proteins). (Slide 20)

10. Nucleosomes both protect DNA and allow it to be packaged; this in turn allows DNA to be supercoiled. (pg. 91)

a. Review (slides 20-23) and briefly outline why it is essential to supercoil chromosomes.

b. Outline how nucleosomes help regulate transcription. (Slide 19)

11. Which stage of Mitosis would be supercoiled begin? (Slide 22)

7.1 S.2 Utilization of molecular visualization software to analyse the association between protein and DNA within a nucleosome.

12. Use the RCSB Protein Bank to read about nucleosomes and examine Jmol images of them.Article on nucleosomes: http://www.rcsb.org/pdb/101/motm.do?momID=7Jmol visualisation of a nucleosome: http://www.rcsb.org/pdb/explore/jmol.do?structureId=1AOI&bionumber=1

Page 7: Topic Five: Nucleic Acids student notes

Topic Five Nucleic Acids Name:

a. Identify the two copies of each histone protein. This can be done by locating the ‘tail of each protein’. The tails of the proteins are involved in regulating gene expression.

b. Suggest how the positive charges help to form the nucleosome with the negatively charged DNA molecule.

c.

Review question

13. Which of the following is the end product of DNA replication in a human somatic cell?a. 23 chromosomesb. 46 chromosomesc. 23 pairs of chromosomesd. 23 pairs of sister chromatids

2.7 U.2 Helicase unwinds the double helix and separates the two strands by breaking hydrogen bonds. 2.7 U.3 DNA polymerase links nucleotides together to form a new strand, using the pre-existing strand as a template.

14. Label the diagram below. (Slide 25)

2.7 U.1 The replication of DNA is semi-conservative and depends on complementary base pairing .

15. Explain how complementary base pairs are arranged. (Slide 28)

16. Explain why DNA replication of this kind of referred to as being semi-conservative. (Slide 28)

2.7 A.1 Use of Taq DNA polymerase to produce multiple copies of DNA rapidly by the polymerase chain reaction (PCR).

17. Polymerase chain reaction (PCR) is a key technique in DNA manipulation and analysis.a. State the main uses of PCR. (Slide 30)

Page 8: Topic Five: Nucleic Acids student notes

Topic Five Nucleic Acids Name:

b. Complete outline of the three keys steps in the process of PCR. (Slide 30) Denaturation:

Annealing:

Elongation:

c. If one cycle of PCR yields two identical copies of the DNA sequence. Calculate how many copies 20 cycles would yield.

2.7 S.2 Analysis of Meselson and Stahl’s results to obtain support for the theory of semi-conservative replication of DNA. (Slides 31-32)

The image to the right details the three possible methods of DNA replication.

Review your understanding of Meselson and Stahl’s experiments by using using the presentation, the McGraw and Hill animation (clear and accessible) or the Scitable article (Nature education) (more detailed).

17. Describe and explain the result found by centrifuging a mixture of DNA from generation 0 and 2. (2)

7.1 U.3 DNA polymerases can only add nucleotides to the 3’ end of a primer. (Slide 33)

18. Outline what a primer is and the role it has in DNA Replication.

19. In which direction does DNA polymerase move along the template strand? What implication does this have for the addition of bases on the growing strand?

7.1 U.4 DNA replication is continuous on the leading strand and discontinuous on the lagging strand. 7.1 U.5 DNA replication is carried out by a complex system of enzymes.

https:// upload.wikimedia.org /wikipedia/commons/a/a2/DNAreplicationModes.png

Page 9: Topic Five: Nucleic Acids student notes

Topic Five Nucleic Acids Name:

20. Explain the process of DNA Replication:a. Distinguish between the lead strand and the lagging strand. (slide 34)

21.a. Summarize the roles of the enzymes of DNA Replication: (Slide 37)

Enzyme Function

DNA Gyrase(aka topoisomerase)

an enzyme that relieves strain while double-strand DNA is being unwound by helicase

DNA Helicase

DNA Polymerase III

RNA Primase

DNA Polymerase I replaces RNA primers with DNA

DNA Ligase

22. Construct a replication fork containing all the enzymes listed above.

7.1 A.2 Use of nucleotides containing deoxyribonucleic acid to stop DNA replication in preparation of samples for base sequencing. (slide 47)

24. State how dideoxyribonucleic acid affect DNA replication.

Page 10: Topic Five: Nucleic Acids student notes

Topic Five Nucleic Acids Name:

25. State what is attached to deoxyribonucleic acid during base sequencing.

2.7 U.4 Transcription is the synthesis of mRNA copied from the DNA base sequences by RNA polymerase. 2.7 U.5 Translation is the synthesis of polypeptides on ribosomes

26. What are the names of the two processes required for gene expression? (Slide 49)

27. Give 3 examples of proteins constructed using the two processes in question 22 and their functions (Slides 49-54)

a.

b.

c,

28. Where do transcription and translation take place in the cell? (Slide 55)

7.2 U.2 Nucleosomes help to regulate transcription in eukaryotes. (Slide 56)

29. State the name of the key chemical group that can be added to DNA to affect transcription.

30. Outline how the modification of DNA and histones can activate or deactivate genes.

31. Outline gene expression of the length of your nose. How might changes in the production of the molecules morphogens effect nose length? (Slide 57)

32. Suggest how modified DNA or histones maybe inherited by offspring (Hyperlink on Slide 58).

Page 11: Topic Five: Nucleic Acids student notes

Topic Five Nucleic Acids Name:

33. State the name of the branch of genetics concerned with these heritable changes not caused by DNA. (Hyperlink on Slide 58)

7.2 U.6 The environment of a cell and of an organism has an impact on gene expression. (Hyperlink on Slide 58)

34. Give an example of how the environment of an organism impacts gene expression.

7.2 S.1 Analysis of changes in the DNA methylation patterns. (Slide 59)

35. How does methylation effect gene expression?

2.7 U.4 Transcription is the synthesis of mRNA copied from the DNA base sequences by RNA polymerase.

36. Outline the roles of RNA polymerase and ribonucleoside triphosphates in the process of creating mRNA. (Slide 61)

7.2 A.1 The promoter as an example of non-coding DNA with a function.(Slides 63 & 64)

37. Coding regions are used as a guide for the production of polypeptides, but non-coding regions are not. Non-coding regions do however have important functions, for example promoters. Outline how promoter regions called the TATA box of DNA molecules aid the production of polypeptides.

7.2 U.1Transcription occurs in a 5’ to 3’ direction.

38. State the direction of transcription and draw a simple diagram to show the addition of an RNA molecule to a growing mRNA strand. (Hyperlink on Slide 65)

7.2 U.3 Eukaryotic cells modify mRNA after transcription.

39. Distinguish between introns and exons in eukaryotic DNA. (Slide 66)

http://www.phschool.com

Page 12: Topic Five: Nucleic Acids student notes

Topic Five Nucleic Acids Name:

40. Most of the eukaryotic genome is non-coding. There are two types of repetitive sequences: moderately repetitive sequences and highly repetitive sequences otherwise known as satellite DNA. Give an example of a region of DNA that contains highly repetitive sequences and outline the function of that region. (Slide 67)

7.2 U.4 Splicing of mRNA increases the number of different proteins an organism can produce.(Slide 70)

41. The splicing process above can happen in different ways to the same gene.a. Describe how this happens.

b. Name an example of a protein family that is commonly alternatively spliced.

7.1 A.3 Tandem repeats are used in DNA profiling. (Slide 71 & 72)

42. State the two different sources of DNA used in paternal and maternal profiling.

43. Suggest a reason why non-coding regions are more useful than coding regions in DNA profiling.

44. Describe what is meant by the term tandem repeat sequence.

7.2 U.5 Gene expression is regulated by proteins that bind to specific base sequences in DNA.

45. One well known example of the regulation of gene expression by proteins is the metabolism of lactose in E. Coli (prokaryotic bacteria). The diagram below illustrates this example. Complete the cloze to outline this example of gene expression. (Slide 73-76)

Page 13: Topic Five: Nucleic Acids student notes

Topic Five Nucleic Acids Name:

http://commons.wikimedia.org/wiki/File:Lac_Operon.svg

_____________ (2) protein binds to the

operator. _____________ (1) cannot bind

to the ___________ (3) therefore the

genes that produce proteins involved in

lactose metabolism is not transcribed.

____________(5) binds to the repressor.

The ____________(2) cannot bind to the

operator. RNA polymerase binds to the

_____________ allowing the genes that

produce proteins involved in lactose

metabolism ___________ be transcribed.

Key1 RNA Polymerase 4 Operator2 Repressor 5 Lactose3 Promoter 6, 7 & 8 Genes that produce proteins (e.g. enzymes involved in lactose metabolism)

46. Complete the table to outline the functions of the different types of gene expression regulation by proteins that occur in eukaryotes.(slide 77)

DNA Sequence Function

Enhancers

Silencers

A region of DNA located close to a specific gene. Once bound to the sequence RNA polymerase transcribes the gene.

Slide 75

Page 14: Topic Five: Nucleic Acids student notes

Topic Five Nucleic Acids Name:

Nature of Science: Looking for patterns, trends and discrepancies - there is mounting evidence that the environment can trigger heritable changes in epigenetic factors. (3.1)

47. DNA and Genetics is fast evolving area of Biological research. One of the very interesting areas of current research is Epigenetics. Read the article about Epigenetics on Learn.Genetics (http://learn.genetics.utah.edu/content/epigenetics/).

a. Outline what Epigenetics is a study of.

b. Discuss how Epigenetics both fits into the accepted understanding of DNA expression and how is contrary to it.

7.2 U.6 The environment of a cell and of an organism has an impact on gene expression.

48. Outline how temperature effect gene expression Artic Hares. (Slide 78)

7.3 S.2 The use of molecular visualization software to analyse the structure of eukaryotic ribosomes and a tRNA molecule.

49. Draw a two diagrams to outline the structure of a ribosome. Label the following: (Slide 82 &83)a. large subunitb. small subunitc. three rRNA binding sites (located on the large subunit)

50. Label the following regions on this generalized tRNA molecule, outlining the function of each. (Slide 84)

Page 15: Topic Five: Nucleic Acids student notes

Topic Five Nucleic Acids Name:

7.3 A.1 tRNA-activating enzymes illustrate enzyme–substrate specificity and the role of phosphorylation.

51. What role does ATP playing the constructing a tRNA? (Slide 86)

7.3 U.1 Initiation of translation involves assembly of the components that carry out the process. AND 7.3 U.2 Synthesis of the polypeptide involves a repeated cycle of events. AND 7.3 U.3 Disassembly of the components follows termination of translation.

52. Complete the table to summarize the process of translation. (Slide 90)

a. Initiation

b. Elongation

c. Termination

2.7 U.6 The amino acid sequence of polypeptides is determined by mRNA according to the genetic code. 2.7 U.7 Codons of three bases on mRNA correspond to one amino acid in a polypeptide .

53. Define mRNA in terms of it’s function(slide 91)

54. Suggest why the length of mRNA molecules varies. (slide 91)

55. Define the term codon. (slide 92)

2.7 U.8 Translation depends on complementary base pairing between codons on mRNA and anticodons on tRNA. (Slides 93-97)

Page 16: Topic Five: Nucleic Acids student notes

Topic Five Nucleic Acids Name:

56. State the molecule on which anti-codons, which are complementary to codons, can be found.

57. Sketch diagrams below to show translation and use the steps above as annotations.

Page 17: Topic Five: Nucleic Acids student notes

Topic Five Nucleic Acids Name:

2.7 S.1 Use a table of the genetic code to deduce which codon(s) corresponds to which amino acid. 2.7 S.3 Use a table of mRNA codons and their corresponding amino acids to deduce the sequence of amino acids coded by a short mRNA strand of known base sequence.2.7 S.4 Deducing the DNA base sequence for the mRNA strand.

Use the genetic code table to help answer the questions below.

58. Deduce the codon(s) that translate for Aspartate.

59. If mRNA contains the base sequence CUGACUAGGUCCGGAa. deduce the amino acid sequence of the polypeptide translated.

b. deduce the base sequence of the DNA antisense strand from which the mRNA was transcribed.

c. If mRNA contains the base sequence ACUAAC deduce the base sequence of the DNA sense

strand.

60. Transcribe and translate this DNA sequence.

DNA T A C G G G C C C G T G A C A G C C A C T

mRNA

Amino acid

61. An mRNA strand has 76 codons. How many amino acids will be in the polypeptide?

The genetic code – how mRNA codons translate to amino acids

http://www.ib.bioninja.com.au/_Media/genetic_code.jpeg

Page 18: Topic Five: Nucleic Acids student notes

Topic Five Nucleic Acids Name:

62. A polypeptide contains 103 amino acids. What is the length of the gene (unit = base pairs)?

63. A gene is 105kbp (kilobase pairs). How many amino acids are in the polypeptide?

7.3 U.6 Translation can occur immediately after transcription in prokaryotes due to the absence of a nuclear membrane. (Slides 115 &116)

64. State two reasons why translation can occur immediately after transcription in prokaryotes.

7.3 S.1 Identification of polysomes in electron micrographs of prokaryotes and eukaryotes.(Slides 117-119)

65. A polysome is a structure that consists of multiple ribosomes attached to a single mRNA translating it simultaneously to quickly create many copies of the required polypeptide.

a. Label the eukaryote polysome to right to indicate: mRNA Ribosomes Polypeptide chains

b. Describe how polysomes in prokaryotes may differ in structure from polysomes in eukaryotes.

http://www.nobelprize.org/educational/medicine/dna/a/translation/pics_em/polysome.gif

Page 19: Topic Five: Nucleic Acids student notes

Topic Five Nucleic Acids Name:

7.3 U.7 The sequence and number of amino acids in the polypeptide is the primary structure. 7.3 U.8 The secondary structure is the formation of alpha helices and beta pleated sheets stabilized by hydrogen bonding.7.3 U.9 The tertiary structure is the further folding of the polypeptide stabilized by interactions between R groups. 7.3 U.10 The quaternary structure exists in proteins with more than one polypeptide chain. n.b. the next two questions are based on a review of 2.4.U5 and 2.4.U6 and the extension question posed

66. Once translated the polypeptide will naturally fold into a structure. The structure is the result of the polar nature of water in the cytoplasm, hydrogen bonds and interactions between the R-groups.

a. Complete the table to outline the four different levels of protein structure. (slides 120-127)

Notes Fibrous or Globular

Primary (polypeptide)

• The order / sequence of the amino acids of which the

protein is composed

• Formed by covalent peptide bonds between adjacent amino

acids

• Controls all subsequent levels of structure

Neither (– will fold to

become one of the

subsequent levels of

structure)

Secondary

Tertiary

Quaternary