Cycle 3: Inheritance of Sameness Flashcards
How is DNA organized in cell nuclei over the course of the cell cycle?
What do we know about how our DNA is organized in cell nuclei over the course of the cell cycle?
In G1, DNA is organized in pairs of homologous chromosomes. Each chromosome is single stranded.
In S phase, DNA replicates. Now each chromosome is double-stranded.
From S phase to the beginnings of anaphase, chromosomes are still present as homologous pairs of double-stranded chromosomes.
After anaphase, DNA is back to being organized as single stranded homologous pairs
In the G0 phase, the cell is performing maintenance and other functions. Some cells never leave the G0 cycle. What kinds of cells do you think that may be? Cells that rarely divide, so like nerve cells or mature cardiac (heart) cells. Some cells don’t even go into G0 really at all because they are on the grindset of dividing.
How is DNA organized in cell nuclei over the course of meiosis?
What do we know about how our DNA is organized in cell nuclei over the course of meiosis?
In G1, DNA is organized in pairs of homologous chromosomes. Each chromosome is single stranded.
In S phase, DNA replicates. Now each chromosome is double-stranded.
From S phase to the beginnings of anaphase I, chromosomes are still present as homologous pairs of double-stranded chromosomes.
During anaphase I, homologous pairs are separated and the genome size is reduced to half. Each cell is haploid.
During anaphase II, sister chromatids are separated
Difference between n and C
“n” identifies how many nuclear chromosomes are unique.
Coefficient of n identifies how many sets of n there are.
Notes:
1) Total number of chromosomes = number of centromeres
2) N does NOT change during mitosis
3) Coefficient of n is also known as PLOIDY
Most animals are 2n/diploid.
As humans, only 23 of our chromosomes encode everything we need to grow and live, but we have 2 copies of each chromosome. Because we inherit one chromosome from mum and one chromosome from dad.
“C” = amount of DNA (pg units or base pairs) for 1n
Coefficient of C = number of copies of the entire genome
Notes:
1) The word “genome” is synonymous with 1C !
2) Tip: the coefficient of C can be found by counting the number of chromatids per homologous grouping.
3) The value of C cannot be determined
just by looking at a karyotype.
C = amount of DNA in one set → in one set we have a “c” amount of DNA → in humans, we have 2 sets so our C coefficient is 2C in the G1 phase
When DNA is replicated, which of n and C changes?
Only the coefficient of C changes (1C → 2C)
How do we know?
In the first picture we have 2 centromeres, representing 2 homologous chromosomes
In the second picture WE STILL HAVE 2 CHROMOSOMES, because we still only have 2 centromeres
Therefore, the ploidy has not changed and n has not changed. These values would only change if we somehow created new chromosome or lost chromosomes, and this does not happen.
Ex. in humans:
N = 23 chromosomes
We are diploid (2n). Remember that we have 46 chromosomes and are 2n, with each set coming from our biological parents.
Before replication 2N = 46 chromosome
After replication, we still have 46 chromosomes (not 92 chromosomes). We are still human. Meaning we still have 2n.
After replication, we have 92 chromatids
In the process of DNA replication, enzymes string DNA nucleotides together to form more DNA. So there is an increase in the total amount of DNA. But is there a difference in how much DNA constitutes our genome? No. Our genome is 1C and that has not changed. DNA replication only changes how many copies of our entire complement of DNA is in the nucleus.
In questions on exams, make sure you note if the question is asking you about the n-value, coefficient of n/ploidy, C-value, or coefficient of C/copies of genome.
-values and C-values have to be characterized independently! It is not reasonable to assume that a larger or smaller n-value also corresponds with a larger or smaller C-value, respectively.
Sometime, a large amount of DNA could be packed into only a few chromosomes, or conversely, a small amount of DNA could be packed into a large number of chromosomes. The size of chromosomes is not constant–– not within one organism nor between organisms.
DNA
Made up of 4 nucleotides:
PURINES = Adenine (A) and Guanine (G)
PYRIMIDINES = Thymine (T) and Cytosine (C )
Base-pairings
A-T ; 2 H-bonds
C-G ; 3 H-bonds
Nucleotides which form the sugar-phosphate backbone of DNA each contain one of 4 nitrogenous bases: A, G, T, C
A + G = purines (2 carbon rings)
T + C = pyrimidines (1 carbon ring)
Nucleotides form complementary base-pairs: a purine must pair with a pyrimidine
A pairs with T
G pairs with C
Base pairs are stabilized by hydrogen bonds
Phosphodiester bonds link nucleotides together
DNA has a direction
5’ end → phosphate group
3’ end → OH group
DNA runs antiparallel
The phosphate group of new nucleotides must be added to a free 3’ OH (i.e. replication occurs in the 5’ to 3’ direction)
DNA is antiparallel, meaning the 3’ end of one strand is opposite the 5’ end of its complementary strand
DNA Replication
DNA Replication is Semi-Conservative
Helicase unwinds the double helix to separate the 2 parent strands
Parent strand becomes a template for synthesis of a new strand
Happens during S phase before
mitosis
Replication begins with helicase unwinding DNA at the ori by breaking H-bonds between complementary base pairs
SSBs stabilize the strands from snapping back together
Topoisomerase unwinds DNA further downstream to relieve the DNA from the stress of helicase unwinding it
Each parent strand becomes a template for synthesis of a new strand
Semiconservative replication → Resulting DNA contains one “old” strand from the original DNA molecule and one newly synthesized strand
DNA polymerase
Polymerase III extends primers by adding nucleotides to form new DNA complementary to template strand
Synthesizes new strand 5’-3’
Reads template strand 3’-5’
Polymerase I cleaves off RNA primer and replaces it with DNA nucleotides
5’-3’ exonuclease activity
U –> T
DNA POLYMERASE III:
DNA polymerase III can only add bases to the free 3’ hydroxyl group
New strand = synthesized in 5’-3’ direction
Template strand = read in 3’-5’ direction (because DNA is anti-parallel)
DNA POLYMERASE I:
Removes RNA primers and replaces with DNA, synthesizing in 5’-3’ direction
U –> T
Continuous vs. Discontinuous
Leading strand = continuous replication towards the replication fork
Lagging strand = discontinuous replication away from replication fork (Okazaki fragments)
** Remember that both leading and lagging strands begin with an RNA primer!
However, only the lagging strands have okazaki fragments.
Leading strand synthesized continuously towards replication fork as helicase unwinds DNA
Lagging strand synthesized discontinuously away from replication fork, forming Okazaki fragments
Ligase seals nick between Okazaki fragments
Replication Bubbles
Multiple origins (bubbles) increase replication efficiency
Each replication origin has 2 replication forks travelling in opposite directions
Any one particular strand of DNA is replicated continuously at one fork but discontinuously at the other fork
While prokaryotes have a single replication bubble, eukaryotes have multiple origins of replication to increase the speed at which at which our DNA is replicated because our chromosomes are longer
· Each replication bubble has 2 replication forks travelling in opposite directions
o Eventually, the replication forks of different replication bubbles meet and the chromosome will be fully replicated
· Each strand in the replication bubble is replicated continuously at one fork but discontinuously at the other fork
· Pay attention to if the question wants you to identify the first couple of bases because that would be the RNA primer, so U instead
Cell Senescence
Telomeres are repeating TTAGGG sequences at the ends of your chromosomes
They act as buffer regions
Hayflick limit: the number of times a somatic cell can divide before reaching the end of the TTAGGG region.
Cell senescence: irreversible cell cycle arrest
Telomerase
DNA replication leaves the 5’ end of newly-synthesized strands shorter than their complementary strand
Telomerase restores the length of chromosomes (but does not prevent DNA shortening!)
Telomerase binds to the 3’ end of the template strand and extends the length of the template strand using a short RNA template built into the enzyme.
Consider the primer closest to the 5’ end of a chromosome. That primer is RNA, so it must be removed by DNA polymerase I and replaced with DNA nucleotides. However, in order for new nucleotides to be added, DNA polymerase III requires a free 3’ end. It has no 3’ end since there are no more bases upstream. Therefore, the RNA primer cannot be replaced.
This is ok. As long as there are more telomeres (TTAGGG), there is no harm done by removing a couple bases. Only when the telomere buffers have all been depleted is there a risk that DNA shortening will cut into essential genes.
RNA primer synthesized on newly extended template strand
DNA polymerase III extends the primer using complementary base pairing
Remember: in which cells would telomerase be active?
Complementary RNA primer can now be added onto template strand
DNA polymerase III extends the primer using complementary base pairing - telomere is now a double stranded TTAGGG sequence!
Telomerase only active in stem cells, germ cells, cancer cells, etc. (not active in somatic cells)
