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Summary PCB 3063 General genetics Exam 4 Study Guide

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PCB 3063 General genetics Exam 4 Study Guide (summary)

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PCB 3063 Exam 4 Content

Content: Chapter 9, 11, 12, and 13

Quizlets:
● Ch. 9 Homework: https://quizlet.com/_alimez?x=1jqt&i=ss3eh
● Ch. 11 Homework: https://quizlet.com/_am85g3?x=1jqt&i=1kknp8
● Ch. 12 and 13 Homework: https://quizlet.com/_am7vkb?x=1jqt&i=ss3eh

Chapter 9: Molecular Structure of DNA and RNA
1. What are nucleic acids?
DNA or RNA; large macromolecules composed of repeating nucleotides.
● They form the repeating units
● The nucleotides are linked to form a strand
● Two strands can interact to form a double helix
● The double helix folds, bends and interacts with proteins resulting in 3-D structures in the
form of chromosomes.
Note: Genetic Material must fulfill 4 criteria:
1) Contains information necessary to construct an organism (provides the blueprint for
determining the inherited traits of an organism.
2) Transmission from parents to offspring through reproduction
3) Replicates due to genetic material being passed from parents to offspring, and cell
division
4) Variation. It must vary in ways that can account for the known phenotypic differences
within each species.

2. What are nucleotides? What are they composed of? What is nucleoside? Be able to recognize
each component.
a. Nucleotides = repeating structure of nucleic acids (DNA and RNA) linked together in a
linear manner to form a strand. Composed of 1 pentose sugar, 1-3 phosphates, and 1
nitrogenous base. Phosphate (1+) + nucleoside = nucleotide
i. Sugar will be deoxyribose in DNA; ribose in RNA
ii. STRAND = DNA/RNA, nucleotides covalently linked together to form a long, linear
polymer
b. Nucleosides = when a sugar is only attached to a base; NO PHOSPHATE GROUPS ARE
PRESENT

3. What is the composition of a chromosome?
Contains folded/bended DNA interacting with histones (protein)

4. Be able to recognize purines and pyrimidines and how to distinguish them.
a. Purine = double ring structure; Adenine (A) or Guanine (G)
Easy way to remember PUR As Gold

, b. Pyrimidine = single ring structure; Thymine (T), Cytosine (C) , Uracil (U)
i. U only found in RNA
Easy way to remember CUT the Py




5. Given a strand of DNA or RNA, be able to determine the 5’ to 3’ orientation. In linear DNA,
what are functional groups you expect at the free 5’ and 3’ ends?

Ex: DNA molecule with sequence 5’-TTCCGGAT-3’

● DNA pair: 3’-AAGGCCTA-5’
● RNA from original: 3’-AAGGCCUA-5’
● RNA pair from new strand: 5’-UUCCGGAU-3’

Functional groups at the free ends:

● 5’ end → phosphate groups
● 3’ end → OH



6. How are adjacent nucleotides linked to
each other?
a. Nucleotides are covalently bonded together
by phosphodiester bonds
i. A phosphate group connects two
sugars via ester bonds -> phosphodiester linkage
in DNA and RNA where phosphate links two
sugars at 3’ by ester bonds → phosphates and
sugars form the backbone portion of DNA/RNA strand; composed of the repeated covalent
linkage of the phosphates and sugar molecules
ii. phosphodiester linkage: in a DNA or RNA strand, a linkage in which a phosphate group
connects two sugar molecules.

, iii. Backbone: the portion of a DNA or RNA strand that is composed of covalently linked
phosphates and sugar molecules.
b. Peptide bonds connect amino acids

7. Chargaff’s rules of base composition
The observation that in DNA:
a. the amounts of A and T are equal; as are the amounts of G and C (A=T; G=C)
b. Purines = pyrimidines or A+G = T+C
c. A+T almost NEVER = C+G


8. Given some base compositions, determine remaining base composition.
a. ​Consider a piece of double-stranded DNA in which each strand of the helix has 1,800
nucleotides. If the piece of double-stranded DNA had a total of exactly 600 deoxycytidine
monophosphate nucleotides, how many deoxyguanosine monophosphate nucleotides would
be present? [When a nucleoside triphosphate is incorporated into DNA, it loses two
phosphates to form a monophosphate.]
Answer: 600

(1800-(600*2)=600

b. Consider a piece of double-stranded DNA that is 2,000 base pairs in length. If this piece of
DNA had a total of 600 deoxyadenosine monophosphate nucleotides, how many
deoxycytidine monophosphate nucleotides would be present?
Answer: 1400

2000x2= 4000 nucleotides. 4000 -1200 (600A +600T)= 2800/2= 1400 C, 1400 G
9. History of DNA discovery roles of Franklin, Wilkins, Watson and Crick and the approaches and
techniques they used
a. Wilkins & Franklin
Their experiment can be summarized as such:
● A DNA fiber is exposed to X-Rays
● The scatter of rays from the fiber is caught on a photographic film
● The uniformity or irregularity of the scatter is indicative of the structure of DNA
i. Used x-ray diffraction data to produce a well defined diffraction pattern if molecular
structure has a normal pattern
ii. Extracted finer DNA fibers than ever before
iii. Found: DNA consistent with helical structure, diameter was too wide to be a single
helix, about 10 bp per complete turn, 20A thickness with groups spaced 3.4 A (0.34nm)
along fiber & repeating units every 34A (3.4nm)
b. Watson & Crick
They did not perform any experiments
i. Assumption: DNA nucleotides are positioned linearly, chemical linkage between
two nucleotides is always the same

, ii. Produced even clearer x-ray diffraction image
iii. Ball and stick model showed: A=T, G=C, double helix

10. Structure of double helix proposed by Watson and Crick (strand directionality, base pairing,
number of base pairs per twist, handedness of helix, location of sugar phosphate backbone,
location of the bases, etc)
a. Key Features:
i. 2 strands of DNA form a right-handed, clockwise double helix and are twisted
around a common axis
ii. The bases in opposite strands hydrogen bond according to the AT/GC rule
iii. The 2 strands are antiparallel with regard to their 5’-3’ directionality
iv. There are ~10 nucleotides (bp) in each strand per complete turn of the helix
v. Sugar-phosphate backbone on the outside and the nitrogenous bases on the inside
b. One complete turn = 3.4 nm, one nucleotide = 0.34 nm, sugar phosphate backbone is located on
outside of the helix; based are located on inside

11. What is the width of a common form of DNA?
a. 2 nm

12. Spacing of the adjacent nucleotide pairs in common form of DNA in Angstroms
The distance between two DNA strands is 20 Angstroms (2nm)
Each turn of DNA is made up of 10.4 nucleotide pairs and the center-to-center distance between
adjacent nucleotide pairs is 34 Angstroms (3.4nm).
13. Hydrogen bonding between A & T and G & C (in each case be able to recognize H-bond donor
and acceptor)
a. Two hydrogen bonds between A&T and three hydrogen bonds between G&C.
i. G&C more stable; there are more sites to hydrogen bond to giving rise to the
triple bond = more stable
ii. A&T are less stable and easier to break apart during replication

14. Features of major and minor groove, be able to trace them if given a diagram. Their widths?
a. “Grooves” = indentations where h2o surrounds atoms of the bases. From textbook: the
indentations where the atoms of the bases are in contact with the water in the surrounding
cellular fluid.
i. Minor groove = narrow indentation; bases have access to h2o (1.2 nm)
ii. Major groove = wide indentation; bases have access to h2o (2.2 nm)
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