Molecular Biology Exam 1, Genes are DNA PT 1

Chromosomes

• physical basis of inheritance

• most of the genetic material found with in the nucleus (some in mitochondria)

• genetic material divided into units of densely packed DNA

What are the levels of organization of chromosomes in smallest to largest

nucleosome<30 nm fiber<protein scaffold

nucleosome

• negative DNA wrapped around a positive histone

• DNA + protein (histone)

• 4 subunits of histone

• 200 bp of DNA can be spooled around a single histone

• “molecular spools”

• acetylation - reverses the + charge of the histone to release the DNA, so that the genetic code can be accessed

  

30 nm fiber

• 6 nucleosomes connected together

• involves linker histones

  

protein scaffold

• 30 nm fiber, centromere

• solenoids that are further compacted

Why was it thought that protein was the most likely candidate for the transmission of traits (molecule of inheritance)

diversity of options due to 20 amino acid compounds (compared to the 4 nucleic acids)

Frederick Griffith

• foundational study in discovering DNA as molecule of inheritance

• studied Streptococcus pneumoniae, which infections are usually lethal in mice

• observed that certain colonies were less virulent and that there were morphological differences as well (virulent = smooth, polysaccharide capsule) (nonvirulent = rough)

experimental design of Griffith’s experiment

• step 1: inject virulent (S) into the mouse = death, inject nonvirulent strain = mouse survives

• step 2: boiling of S strain + injecteion of heat killed S cells into mouse = mouse survive, boiling S strain + inject heat killed S cell + live R strain = death

In Griffiths experiments, why did the mouse injected with the heat killed S strain (virulent) + live R strain (nonvirulent) = death?

• the cellular debris from the S strain was able to convert the R strain into the virulent form via transformation

• the gene causing the protective polysaccharide capsule is transformed from the S strain to the R strain, causing the R stain to become indetectable by the mouses immune system, resulting in death

who did an experiment based on the info gathered in Griffiths experiment to determine the molecule of inheritance

Avery, Macleod, & McCarty

Avery, Macleod, & McCarty experimental design

• RNase added to the heat killed sample → breakdown of RNA → injected into mouse → DEATH

• protease added to heat killed sample → breakdown of proteins → inject into mouse → DEATH

• DNA added to the head killed sample → breakdown of DNA → inject into mouse → SURVIVES

What two experiments strongly implicated that DNA was the molecule carrying genetic info

Griffith’s and Avery, Macleod, & McCarty’s

Hersey- Chase Experiment

• bacteriophage and E. coli

• the components of DNA were labeled with different biomarkers

  • phosphorous 32 P - found in DNA, and not proteins

  • sulfur 35 S - found in proteins but not DNA

• two separate cultures of E coli were infected and then agitated to cause dissociation of phage and bacteria

• In 32 P labeled phages → most of the radioactivity was associated with the bacterial cell fraction (DNA entered the host cell = DNA was the molecule of inheritance )

• In 35 S labeled phages → most of the radioactivity associated with phage ghosts fraction (protein did not enter the host cell)

• the “capstone” experiment that proved DNA as the molecular of inheritance

what three key properties were assumed for DNA

1. DNA must allow for successful replication of the genetic material at every cell division (“high fidelity”, structure allows for fateful replication)

2. DNA must encode all the information needed for the assembly of proteins by an organism (proteins)

3. DNA must be able to change (mutations must be possible)

*also, structure must be relatively stable as so that organisms can rely on its ended information

what information was already known prior to Watson and Crick’s discovery of the double helix

1. building blocks (4 nucleotides)

2. chargarff’s base composition rules

3. x-ray diffraction data (double helix)

what are the 3 chemical components of DNA

• Phosphate, deoxyribose (sugar), nitrogenous base (adenine, guanine, cytosine, thymine)

• phosphate → allows linkage

purine nucleotides

adenine, guanine; double ring structures

pyrimidine nucleotides

cytosine, thymine, uracil (RNA only); single ring structures

what are the purines matched to the pyrminidines

• A - T (or U)

• G - C

DNA has a — at the 2’ position

H

what are regions with many A-Ts called?

the TATA box, which is found at the start of the genome and origin of replicaiton. A-T forms 2 Hydrogen bonds, meaning it is more easily accessed than G-C, which has 3 hydrogen bonds.

chargaff’s rules of base composition

1. the total number of pyrimidine nucleotides (T+C) always equals that total number of purine nucleotides (A+G)

2. Amount of T= amount of A; amount of C= amount of G)

Rosalind Franklin, X-ray diffration

• diffraction pattern showed regular, repeating features consistent w'/ a helix = helped to establish dimension of the DNA (diameter and spacing between bases

• showed that there were 10 base pairs per turn

• the distance between bases was 3.4 A

• helical structure allows phosphate backbone to face outwards and protect the bases and H-bonding within

DNA structure

• Watson & Crick → composed of two side by side chains of nucleotides; bases project to the center and pair

• two strands twist into the shape of a double helix; sugar phosphate backbone forms the sides of the helix

  • DNA is typically right handed double helix

• in the analogy of DNA resembling a spiral staircase, think of the hydrogen bonds as “steps”

  • sugar phosphate backbone: sides and rail

  • base pair: steps

  • H-bond - connections holding the steps together

periodicity

• each base pair is twisted from the previous one by 36 degrees

• takes 10 bp to go completely around the helix

• info taken from rosalind frankling diffraction research

major groove and minor groover

• the two distinctive sizes of grooves that are not equal in size

• major groove contains most DNA-protein interactions (for replication, transcription) (does not require opening and closing of sections of DNA when scanning for correct activation sites)

• minor = 12 A

• major = 22 A

• 12 + 22 = 34 A (length of one turn in DNA)

major and minor grooves present a chemical code for protein binding

• major and minor grooves expose different chemical groups of DNA bases

• pattern provide a chemical code that proteins recognize by interacting with the chemical groups

• major groove each bp present a characteristic pattern of:

  • A = hydrogen bond acceptor

  • D = hydrogen bond donor

  • M = nonpolar methyl group

  • H = nonpolar hydrogen

    

nucleotide

phosphate, sugar, and nitrogenous base; the building block of nucleic acids

where do phosphates and new growth attach to on a nucleotide

at 3’

which DNA groove gives enough info for proteins to read the DNA sequence, without having to open up the two separate strands. why?

major; the pattern of hydrogen bonds and methyl groups allow proteins to recognize specific regions of DNA

DNA backbone

• alternating phosphate and deoxyribose sugar units;(sugar phosphate backbone)

• phosphosiester linkage connects the 5’ carbon of one deoxyribose to the 3’ carbon of another deoxyribose

• said to 5’ to 3’ polarity (key, since DNA strands are synthesized in a 5’ to 3’ direction

• phosphate backbones run in an antiparallel direction → adds stability to DNA molecule

which type of pairing between bases is most stable

C-G ; requires higher temp. to separate two strand of DNA