BSC 2010C Lecture 16: CH16 The Molecular Basis of Inheritance

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19 Apr 2018

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Prokaryote vs. Eukaryote organization of DNA

Pro have 1 circular, chromosome made up of double-stranded DNA and a small

amount of protein

Chromosome is supercoiled and resides in the nucleoid of the cell

○

Euk have linear chromosomes made up of double stranded DNA and many

proteins

Eukaryotic organization of DNA

DNA organization:

Linear chromosomes

○

Double stranded

○

Wrapped around protein complexes called histones

DNA + histone = nucleosome

○

DNA and proteins are organized as cylindrical fibers called chromatin

○

Chromosomes are made up of both DNA and protein

DNA provides instructions for synthesizing proteins

Proteins determine organismal biochemistry, anatomy, physiology,

behavior, etc.

○

Nitrogenous Bases

Single Rings (pyrimidines)

Cytosine in DNA & RNA

○

Thymine in DNA

○

Uracil in RNA

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Double Rings (purines)

Adenine in DNA and RNA

○

Guanine in DNA and RNA

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Amino Acid Monomers

All amino acids have an amino group and a carboxyl group

Structurally, amino acids have an alpha carbon in the center. Attached to the

alpha carbons are four things:

Amino group

○

Carboxyl group

○

Hydrogen atom

○

An R group

Side chain group that varies in different amino acids

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Early Understanding of Inheritance

DNA is comprised of nucleotides

Phosphate and sugar (deoxyribose) backbone

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4 different nucleotides

○

Proteins are polymers comprised of amino group, carboxyl, and one of 20

different amino acids

Chargaff's Rules

Showed that nucleotide composition varies between species

Shows that amount of A= amount of T, and amount of C= amount of G

Base Pairing Rules

DNA width is consistent down the entire length of the DNA strand

A and G are double ring purines

T and C are single ring pyrimidines

A and G bond with each other - DNA is too wide

○

T and C bond with each other - DNA is too narrow

○

If purines bind with pyrimidines, DNA width is consistent

In DNA

C -G

A - T

○

In RNA

C - G

A -U

○

DNA strands: anti-parallel and complimentary

The combination and sequence of nucleotides in DNA provides the blueprint for

making proteins that make an organism what it is.

Just four nucleotides account for all the diversity on earth

○

Nucleotides make up the genes that are passed down from one gen to the

○

DNA Replication (synthesis phase)

Semi-conservative

The parent DNA strands unwinds and two new strands are created, using the

base-pairing rules

Result is that the new strands are comprised of one parent strand and one new

strand.

DNA Replication Origins

Replication starts at replication origin

Prokaryotes have only one replication origin

Circular DNA

○

Eukaryotes have many (up to hundreds or thousands) replication origins

Linear DNA

○

Replication proceeds in both directions

DNA Synthesis

DNA double helix in "parent" strand must be opened and unwound

This gives room for new nucleotides to be added to both sides

Helicase is the enzyme that unwinds DNA at the replication forks

Then single-strand binding proteins attach to each side of the unwound DNA to

keep them from rebinding

Unwinding of the double helix causes strain in front of the replication fork

Topoisomerase relieves this strain by breaking, rotating and then rejoining DNA

strands

The unwound segments of DNA strands are now templates used to add new

nucleotides to create two new complementary DNA strands

Primase starts by adding primer nucleotides

Primer is actually RNA nucleotides

Evolution of RNA to DNA

○

DNA Polymerase 1 will eventually replace RNA primer nucleotides with

DNA nucleotides

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Then, DNA Polymerase 3 add new nucleotides, one at a time, to the template

parent strands

500 nucleotides/second in bacteria

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50/ second in humans

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New nucleotides can only be added to the 3' end of the new DNA strand

Therefore, DNA can only grow in the 5' --> 3' direction

The Molecular Basis of Inheritance

Monday, April 9, 2018

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Document Summary

Pro have 1 circular, chromosome made up of double-stranded dna and a small amount of protein. Chromosome is supercoiled and resides in the nucleoid of the cell. Euk have linear chromosomes made up of double stranded dna and many proteins. Dna and proteins are organized as cylindrical fibers called chromatin. Chromosomes are made up of both dna and protein. Proteins determine organismal biochemistry, anatomy, physiology, behavior, etc. All amino acids have an amino group and a carboxyl group. Structurally, amino acids have an alpha carbon in the center. Attached to the alpha carbons are four things: alpha carbons are four things: Side chain group that varies in different amino acids. Proteins are polymers comprised of amino group, carboxyl, and one of 20 different amino acids. Shows that amount of a= amount of t, and amount of c= amount of g. Dna width is consistent down the entire length of the dna strand.

Related Questions

1.Which of the following is not true about DNA? a) it is found both in the nucleus and the cytoplasm b)it is double stranded and the standards are connected by hydrogen bonds between the base pairs c) A purine base always pairs with a pyrimidine base to make up the DNA code d) it is the control center for both cell activity and heredity 2.Which of the following represents accurate base pairing in DNA ?a) Adenine to Adenine b)Adenine to Uracil c)Thymine to Cytosine d)Adenine to Thymine 3. Purines and Pyrimidines are found a) DNA only b)RNA only c) A and B d)In proteins 4.The bond that forms between carbohydrates is called a)peptide bond b) ester bond c) glycoside bond d)never ending bone 5. Which of the following does not dissolve readily in water? a)proteins b) amono acids c) steroids d) carbohydrates 6. The following arrangement of amino acids represents what level of structure ? ABACABDD a)Primary b)Secondary c)Tertiary d)Quaternary

cerisedog238

In addition to identifying the genetic material, the experiments of Avery, MacLeod, and McCarty with different strains of Streptococcus pneumoniae demonstrated that

	DNA is a double helix.
	DNA may be taken up by bacterial cells and be active.
	DNA is present in bacteriophage T2.
	eukaryotic cells may be genetically transformed.
	DNA binds to the cell wall of the bacterium.

In order to show that DNA in cell extracts is responsible for genetic transformation in Streptococcus pneumoniae, important corroborating evidence should indicate that _______ also destroy transforming activity.

	temperatures high enough to kill cells
	enzymes that hydrolyze proteins
	enzymes that hydrolyze DNA
	enzymes that hydrolyze polysaccharides
	enzymes that hydrolyze RNA

Based on what you have learned about the experiments conducted by Griffith and Avery and colleagues with bacteria, which of the following would result in transformation of living R cells?

	Pure RNA from S cells
	Cell-membrane extracts from S cells
	A mixture of pure RNA and protein from S cells
	Cell-free extracts from S cells
	Pure protein from S cells

A-T base pairs in a DNA double helix

	form three hydrogen bonds with each other.
	are more heat-stable than G-C base pairs.
	are of a substantially different length than G-C base pairs.
	are not accessible to DNA binding proteins.
	are chemically distinct from G-C base pairs.

If 23 percent of the bases in a sample of double-stranded DNA are adenine, what percentage of the bases are uracil?

	27
	50
	0
	73
	23

The uniform diameter of the DNA structure provides evidence for

	antiparallel DNA strands.
	a right-handed helix.
	a double-stranded helix.
	3â² end phosphorylation.
	purineâpyrimidine pairing.

If a sequence of one strand of DNA is 5â²-TGACTATC-3â², what is the complementary strand?

	5â²-ACTGATAG-3â²
	3â²-TGACTATC-5â²
	3â²-ACTGATAG-5â²
	3â²-ACTGATAG-3â²
	5â²-TGACTATC-3â²

What structural aspect of the DNA facilitates dissociation of the two DNA strands for replication?

	3â² end phosphorylation
	Purineâpyrimidine pairing
	Antiparallel DNA strands
	Covalent bonds between sugars and phosphate groups
	Hydrogen bonding between paired bases

If the MeselsonâStahl density gradient experiment had resulted in two bands of DNA molecules after only one round of replication, one containing only ¹⁵N and the second only ¹⁴N, this result would have indicated that replication was

	unidirectional.
	dispersive.
	conservative.
	bidirectional.
	semiconservative.

10.

The nucleoside analogue acyclovir, which is used to treat herpes simplex virus (HSV) infections, lacks a 3â² hydroxyl group (âOH). Predict what will happen if the host cell DNA polymerase incorporates a molecule of acyclovir into an elongating strand of HSV DNA.

	The replication complex will no longer bind to origins of replication.
	The DNA polymerase will only incorporate acyclovir molecules.
	Acyclovir will bind single-strand binding proteins.
	DNA polymerase will not be able to link a successive nucleotide.
	Acyclovir will block the activity of the DNA helicase.

11.

Which of the following does not demonstrate the stability of the DNA double helix?

	Single-strand binding proteins are required to keep the strands apart.
	Paired bases form hydrogen bonds.
	High temperatures are required to denature it.
	DNA synthesis requires energy.
	DNA helicases require ATP.

12.

What effect would a primase inhibitor have on DNA replication?

	DNA polymerase would not be able to add bases to the DNA.
	Primase would be blocked from joining the DNA replication complex.
	Primase would not be able to provide primers for DNA polymerases.
	DNA polymerase would not be able to use DNA as a template.
	There would be no effect on DNA replication.

13.

To replicate their DNA in a timely manner, most eukaryotic chromosomes

	normally have uncoiled DNA.
	have multiple origins of replication.
	contain linear molecules of single-stranded DNA.
	have two replication forks that move in the same direction.
	are composed entirely of DNA.

14.

Which statement about DNA replication is false?

	Okazaki fragments are synthesized as parts of the leading strand.
	Error rates for DNA replication are reduced by proofreading of the DNA polymerase.
	The sliding clamp increases the rate of DNA synthesis.
	Replication forks represent areas of active DNA synthesis on the chromosomes.
	Ligases and polymerases function in the vicinity of replication forks.

15.

In many eukaryotes, there are repetitive sequences called telomeres at the ends of chromosomes. After successive rounds of DNA replication, the _______ strand becomes shorter. In some cells, an enzyme called _______ repairs the shortened strand.

	leading; telomerase
	lagging; telomerase
	lagging; DNA polymerase I
	leading; DNA polymerase I
	leading; replicase

16.

A researcher studies normal human fibroblast cells. They can be maintained in culture but die off after about 30 cell generations. Unexpectedly, a colony of cells continues to survive and divide past 30 generations. Which scenario is most likely true for these cells?

	DNA polymerase is constantly active.
	Primase is able to extend the telomeres.
	The mismatch repair system is extra efficient.
	The telomerase gene is being expressed.
	The cells do not need the ends of their chromosomes.

17.

If DNA polymerase III introduces an incorrect nucleotide, what is the first corrective action made by the DNA repair system?

	The replication complex excises the incorrect nucleotide.
	The excision repair proteins remove the incorrect nucleotide.
	The mismatch repair system scans for base-pairing mismatches.
	There is no correction, because nucleotide errors by DNA polymerases are not repaired.
	DNA ligase connects the correct nucleotide.

18.

Choose the correct order of the following four events in the excision repair of DNA:
(1) Base-paired DNA is made complementary to the template.
(2) Damaged bases are recognized.
(3) DNA ligase seals the new strand to existing DNA.
(4) Part of a single strand is excised.

	1, 2, 3, 4
	2, 1, 3, 4
	3, 4, 2, 1
	2, 4, 1, 3
	4, 2, 3, 1

19.

Six complete cycles of PCR should result in a _______-fold increase in the amount of DNA.

	64
	32
	12
	128
	6

20.

When double-stranded DNA is heated to temperatures above 90Â°C, it denatures. Denaturation is a process that

	breaks covalent bonds.
	hydrolyzes DNA.
	separates double-stranded DNA into two single strands.
	causes new hydrogen bonds to form.
	irreversibly destroys the double helical structure.

1. Which of the following is the correct sequence of events (from start to finish) in cellular fatty acid oxidation?

A. Fatty acid activationÃ carnitine shuttleÃ beta-oxidation

B. Beta-oxidationÃ carnitine shuttleÃ Fatty acid activation

C. Fatty acid activationÃ carnitine shuttleÃ beta-oxidation

D. Carnitine shuttleÃ Fatty acid activationÃ beta-oxidation

2. What reaction is catalyzed by lipoprotein lipase, or lipase?

A. Hydrolysis of TAGs into fatty acids and glycerol

B. Hydrolysis of TAGs into amino acids and glycerol

C. Hydrolysis of cholesterol in to fatty acid and water

D. Hydrolysis of cholesterol into acetyl-CoA

3. How do fatty acids of different carbon lengths cross into the mitochondrial matrix for oxidation?

A. Small FAs of 10 carbons or fewer can freely diffuse through the membrane

B. Large FAs are cleaved into two smaller units that are actively transported across the membrane

C. A membrane-bound transferase catalyzes the conversion of long-chain FAs and FA-carnitine esters, which are transported across the membrane via a specific carrier

D. The breakdown of large FAs starts outside the matrix, allowing for the smaller particles to pass easily through the membrane

4. How can any RNA nucleotide be easily distinguished from a DNA nucleotide?

A. RNA nucleotides have a unique phosphodiester bond linkage

B. RNA nucleotides have uracil nitrogenous bases

C. RNA nucleotides have an -OH on the 2â carbon of the ribose

D. RNA nucleotides cannot be easily distinguished from DNA nucleotides

5. Which of the following statement is true concerning the sequence pTACGTAACGTA?

A. This is an RNA sequence

B. In order to have a more rigid structure in a double-stranded molecule, the sequence given would need more adenosine

C. If the complementary strand were present, the number of purines would be equal to the number of pyrimidines

D. The amount of purines is equal to the amount to pyrimidines

6. Complementary base pairing is considered which level of structure?

A. Secondary

B. Tertiary

C. Quaternary

D. Primary

7. Which of the following is/are required for the polymerization reaction catalyzed by DNA polymerase III?

I. RNA primer

II. DNA primer

III. RNA template

IV. DNA template

V. NTPs

VI. dNTPs

VII. Magnesium

VIII. Calcium

A. II, IV, V, and VII

B. II, IV, VI, and VII

C. I, IV, V, and VIII

D. I, IV, VI, and VII

8. DNA polymerase I, from E. coli, has three different catalytic sites. Which of the following is NOT one of them?

A. RNA-dependent RNA polymerase

B. RNA-dependent DNA polymerase

C. DNA-dependent RNA polymerase

D. DNA-dependent DNA polymerase

9. Eukaryotes solve the problem of replicating the ends of their linear chromosomes by:

A. Using a protein as primer

B. Enzyme activity that uses an RNA template and reverse transcription in a series of repeat cycles

C. All of the listed solution

D. Terminal redundancy whereby each end of the chromosome is duplicated allowing recombination

BSC 2010C Lecture 16: CH16 The Molecular Basis of Inheritance

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