Both RNA and DNA have an OH group on the –
In RNA, the 2′ carbon is –
In DNA, the 2′ carbon is –
Purines are smaller or larger than pyrimidines? –
RNA’s secondary structure results from –
complementary base pairing
The bases of RNA typically form hydrogen bonds with –
complementary bases on the same strand. (it folds back on itself). 3 bonds for g-c, 2 for U-A (no thymine here!)
The RNA strand folds over, forming a hairpin structure: –
the bases on one side of the fold align with an antiparallel RNA segment on the other side of the fold
What structures can RNA molecules have? –
tertiary and quaternary structures
RNA secondary structure –
SINGLE strand on TOP
Double-Stranded region forms a double-helix Nitrogenous bases at the bottom!
RNA vs DNA differences –
1. RNA is a single-stranded molecule
– has a shorter chain of nucleotides.
2. DNA is typically a double-stranded molecule
– has a long chain of nucleotides.
The semi-conservative theory –
can be confirmed by making use of the fact that DNA is made up of nitrogen bases. Nitrogen has an isotope N15 (N14 is the most common isotope) called heavy nitrogen
DNA Gyrase –
Relieves strain of unwinding by DNA helicase; this is a specific type of topisomerase
DNA Ligase –
Re-anneals the semi-conservative strands and joins Okazaki Fragments of the lagging strand
Lengthens telomeric DNA by adding repetitive nucleotide sequences to the ends of eukaryotic chromosomes
DNA polymerase –
DNA polymerase synthesizes a new strand of DNA by extending the 3′ end of an existing nucleotide chain, adding new nucleotides matched to the template strand one at a time via the creation of phosphodiester bonds
Removal of the RNA primer leads to –
the shortening of the chromososme after each round of replication. Chromosome shortening eventually leads to cell death.
An RNA sequence in telemerASE acts as a template for DNA –
This ENZYME adds the telemeric sequence to the 3′ end of the chromosome
Major steps of DNA replication, DNA is opened: –
origin of replication is recognized , two strands are separated by helicase, and replication fork is formed; Single-stranded DNA binding protein (SSB) binds to the single-stranded portion
Major steps of DNA replication, synthesis: –
RNA primer is formed at the beginning of the origin in the 5’to 3′ direction; DNA polymerase begin synthesizing DNA in 5’to 3′ direction, beginning at the 3′ end of each RNA primer; For each replication fork, there is a leading strand and a lagging strand
Major steps of DNA replication, RNA primers are removed by DNA polymerase: – which also fill in the gaps between neighboring fragments ( Okazaki fragments) in the lagging strand
Major steps of DNA replication, Okazaki fragments are linked together by: –
Major steps of DNA replication, DNA Gyrase: –
(topoisomerase) relieves supercoils by transient breaking and resealing one strand of DNA
Ribosomes can be separated into two subunits: –
The small subunit, which holds the mRNA in place during translation. The large subunit, where peptide bonds form
The A site of the ribosome is the –
acceptor site for an aminoacyl tRNA. – charged tRNA
The P site is where –
a peptide bond forms that adds an amino acid to the growing polypeptide chain
The E site is where –
tRNAs no longer bound to an amino acid exit the ribosome
What are bases? –
Either one ring (pyrimidines) or two ring (purines) structures that contain some nitrogen.
By the way, did you know that only Ribose has OH in both the 2′ and 3′ positions? Well..now ya know!!
Only Ribose has OH in what positions? – 2′ and 3′
In order for transcription to begin, RNA POLYMERASE must bind to –
the DNA molecule at the PROMOTOR sequence (The promotor will not be accessable as the DNA is too tightly wound)
the DNA is tightly wound, and transcription canot begin -becuase the promotors cannot be reached by RNA plymerase
UNWOUND DNA in which the promotors are accessable, and Transcription may begin
Heterochromatin – Lies against the nuclear envelope in patches –
and is broken up at the site of the nuclear pore
Topoisomerase 2 –
Type II topoisomerases cut both strands of the DNA helix simultaneously in order to manage DNA tangles and supercoils
Errors in DNA proofreading may result in –
DNA can be broken or altered by various chemicals and types of radiation. For example –
UV light can cause thymine dimers to form, causing a kink in the DNA strand
The nucleotide excision repair system recognizes damage –
Its enzymes then remove the single-stranded DNA in the damaged section
DNA polymerase fills in –
gap caused by damaged strand
DNA ligase seals –
the nick in the repaired strand
After freeing up the bunched up DNA, Topoisomerase does what? –
Defects in the genes required for DNA repair are frequently associated with cancer –
Mutations in the genes involved in the cell cycle go unrepaired, the cell may begin to grow in an uncontrolled manner, which can result in the formation of a tumor
In bacteria, the transcription starting point is SIGMA –
Sigma is in control. It binds to the regulatory region – transcription regulation
Without sigma binding, to the UPSTREAM area –
transcription will not happen
In Eukaryotes, the sigma is called –
The transcription of eukaryotic genes by RNA polymerase generates –
a primary RNA transcript that contains exons and introns.
-Introns are removed by splicing
Small nuclear ribonucleoproteins (snRNPs) form a complex called a spliceosome –
This spliceosome catalyzes the splicing reaction
Are okazaki fragments formed 5 prime to 3 prime? –
Yes! But the lagging strand is READ 3-5 but built 5-3 in sections. This is because DNA synthesis can proceed only in one direction — the 5′ to 3′ direction
The lagging strand in dna replication cannot be a simple mirror image of the leading strand –
It will be going “upstream” instead of “downstream”. The lagging strand therefore has to make lots of little fragments known as okazagi fragments and join them up at the end. It is because of this that lots of RNA primers are used. Rna primers are used to start off each strand of complementary dna, As the lagging strand is unwound, more dna is available for replication and each okazagi fragment is formed
Primase also acts as a halting mechanism to prevent the leading strand from –
outpacing the lagging strand by halting the progression of the replication fork
A mutation is – any permanent change in an organism’s DNA. Point mutations result from a single base change
Mutation types caused by point mutation –
Missense mutation-Result in changes in the amino acid sequence; – a SUBsTITUTION
•Silent mutation-Does not change the amino acid sequence; although one base is changed. no a.a. sequence change
•Nonsense mutation-new codon is a stop codon;
•Frameshift mutation-deletion OR addition of one base;
•Large segment deletion-
•Mutation at splicing sites-
•Triplet repeat expansion-several more copies of a trinucleotide
3 Types of RNA –
1. mRNA=messenger RNA, carrying coding information;
2. tRNA=transfer RNA, carry specific amino acid to ribosome;
3. rRNA=ribosome RNA, form ribosome with ribosomal proteins. Most abundant RNA in a cell
Initiation: In eukaryotes, the small ribosomal subunit binds to the mRNA at –
the 5’cap and slide to the AUG start codon
•Initiator tRNA carrying Met bound to the AUG start codon via base paring with its anticodon.
•Large subunit binds to the small subunit.
•Initially Met-tRNA bind to P-site of the ribosome
•The next tRNA carrying an amino acid binds to A site
Elongation has three steps –
1.Arrival of the aminoacyl tRNA. in the A site
2.Peptide bond formation. amino acids on the tRNAs in the P and A sites
3.Translocation After peptide bond formation, the polypeptide on the tRNA in the P site is transferred to the tRNA in the A site
Translocation occurs when elongation factors move the mRNA down the ribosome three nucleotides at a time, and the tRNA attached to the growing protein moves into the P site
After RNA polymerase binds DNA with the help of other proteins –
it catalyzes the production of an RNA molecule whose base sequence is complementary to the base sequence of the DNA template strand
Eukaryotic genes contain regions called exons and regions called introns –
during RNA processing, the regions coded by introns are removed, and the ends of the RNA receive a cap and tail
Ribosomes translate mRNAs into proteins with the help of intermediary molecules called –
Each transfer RNA carries an amino acid corresponding to the –
tRNA’s three-base-long anticodon
In the ribosome, the tRNA anticodon binds to a three-base-long mRNA codon –
causing the amino acid carried by the transfer RNA to be added to the growing protein
DNA regulatory base sequences (cis regulators) –
Enhancers: regulatory proteins bind to enhancers, transcription begins. Silencers: regulatory proteins bind to silencers, transcription is shut down.
Promoter: upstream , usually within 50 bases from transcription start site DNA regulatory base sequences (cis regulators)
Transcription factors (trans regulators)
activator proteins that bind DNA regulatory base sequences
Turning a gene ON or OFF –
the Acetly group can loosen it UP
The Methyl group PACKS it
Which of the following is a protein that is involved in translation? –
2. Ribosonal RNA
3. RNA polymerase
4. Aminoacyl t-RNA –
3. 1. releases tension, 2. is not a protein
All of the following statements about heterchromatin are true except –
1. Heterochromatin stains more darkly with DNA dyes than does euchromatin
2. Heterochromatin contains more highly condensed DNA than does euchromatin
3. Heterochromatin is associated with inactive genes
4. Heterochromatin is more susceptible to DNasel than is euchromatin –
Euchromatin stains lighter
Heterochromatin is inactive b/c it is too densely packed
The template strand is travelling in what direction? –
3′ to 5′ direction
DNA polymerase can only add in the –
5′ to 3′ direction
OK, think about this…the reason what the lagging strand does not add bases in the same direction is because – originally, it is was part of the DNA strand that is anti-parallel…so it is running in the opposite direction…So it will have to travel from LEFT to RIGHT!!! It adds new bases on the SYNTHSIZED fragments in the 3′ to 5′ direction…The reason why it does not wait for the entire lagging strand to unzip is becuase, it could fold back on itself…get twisted up
DONT FORGET THE RNA PRIMER IS ADDED TO – THE LEADING strand first!!!!!!!!!!!
DNA gets added in the lagging strand by fragments called Okazaki fragments –
DNA polymerase will add new bases in the 5′ to 3′ direction..until it bumps into an RNA primer..in which case it will detach and stop
so..why does it have to add bases in the 5′ to 3′ direction on the leading strand? –
Because…as was said earlier, pyrimidines always pair with purines..that is why it is anti parallel. It needs that primer first! Then it just adds the bases in the direction it would usually go…towards 3
In the lagging strand, the DNA POLYMERASE gets screwed up
becuase – it cannot add bases in the same direction as the leading strand. b/c they are opposite!! Remember it is the DNA polynmerase that is adding the bases..how can it do both at the same time????? Think about it…As it cannot add bases right to left, as is with the leading starnd, it does it from LEFT to right 3′ to 5′ then it has no problems adding the nucleotides. DNA polymerase cannot build a strand in the 3′ → 5′ direction Thus, the strand complementary to the 3′ → 5′ template strand (known as the lagging strand) is synthesized in short segments known as Okazaki fragments. On the lagging strand, primase builds an RNA primer in short bursts. DNA polymerase is then able to use the free 3′ OH group on the RNA primer to synthesize DNA in the 5′ → 3′ direction. The RNA fragments are then removed (different mechanisms are used in eukaryotes and prokaryotes) and new deoxyribonucleotides are added to fill the gaps where the RNA was present. DNA ligase is then able to ligate the deoxyribonucleotides together, completing the synthesis of the lagging strand.
BUT actually…both are still going in the same 5′ to 3′ direction, just REVERSED!!!
The newly synthesized DNA strand is always –
A telomere is a region of repetitive nucleotide sequences at each end of a chromatid –
which protects the end of the chromosome from deterioration or from fusion with neighboring chromosomes
Telomere regions deter the degradation of genes near the ends of chromosomes by –
allowing chromosome ends to shorten, which necessarily occurs during chromosome replication. Without telomeres, the genomes would progressively lose information and be truncated after cell division because the synthesis of Okazaki strands requires RNA primers attaching ahead on the lagging strand. Over time, due to each cell division, the telomere ends become shorter
Telomeres are repetitive sequences at the end of linear eukaryotic chromosomes –
Telomerase replicates the telomere sequences at both ends of a eukaryotic chromosome. Telomerase only exists in germ cells, embryonic cells, stem cells, but not somatic cells
Telemerase solves the END REPLICATION PROBLEM –
Telemerase is BOTH Protein and RNA
Telemerase can reverse transcribe onto its own RNA template to synthesize THE ADDITIONAL DNA. This extends the 3′ telemere. Then polymerase and Okazaki fragments can now extend the 5′ end of the OTHER DNA strand.
There will always be some 3′ SINGLE stranded DNA at the end, but telemereASE will always ensure that our TELEMERES maintain their length and that genes on our chromosomes are protected. The ends of DNA are SINGLE stranded on the –
When a cell is replicating, it can’t copy its DNA all the way to the end –
And so telemeres SHORTEN with each replication. If telomeres get critcally short, the cell cannot function and will die. Picture a shoelace PLASTIC cover that becomes frayed. The ends of CHROMOSOMES have telemeric repeat sequences.
The caps of chromosomes are protected by CAPS called –
telomeres (STEM cells make telomerase)
1. Assures that the chromosome shortening does not occur.
2. Has DNA ploymerase activity
3. Also contains an RNA sequence that provides a template for the syntheses of telemeric repeat DNA
DNA replication – A.K.A. –
RNA synthesis – A.K.A. –
Transcription mRNA (in the nucleus)
DNA synthesis in eukaryote takes place during –
DNA synthesis Required enzymes –
Helicase; ss DNA-binding protein; Primase; DNA polymerase; DNA ligase, DNA topoisomerase II; Telomerase.
DNA repair takes place during G1 phase of eukaryotic cell cycle –
UV radiation causes T-dimers , removed by excinuclease; repaired by DNA polymerase and ligase
Excinuclease – Excision endonuclease, also known as excinuclease or UV-Specific Endonuclease, is a nuclease (enzyme) which excises a fragment of nucleotides during DNA repair. The excinuclease cuts out a fragment by hydrolyzing two phosphodiester bonds, one on either side of the lesion in the DNA. This process is part of “nucleotide excision repair”, a mechanism that can fix specific damages to the DNA in the G1 phase of the eukaryotic cell cycle. Such damages can include the thymine dimers created by UV rays
Which of the following is not required for both DNA replication and RNA transcription (mRNA) –
4. Proteins –
An enzyme that breaks DNA, disperses the tension and releases the strand ahead of a DNA replication growing fork is called a –
2. DNA polymerase
4. Aminoacyl-tRNA synthase –
A lethal mutation occurs in a bacterium, rendering it incapable of replacing its chromosome. Because of this mutation, DNA produces many short fragments of DNA that have RNA sequences at their 5′ ends. The mutation is most likely a gene encoding which of the following? –
1. DNA gyrase
2. DNA helicase
3. DNA ligase
4. DNA polymerase 1
4. DNA polymerase
In bacteria, transcription and translation can occur simultaneously –
Because of limited space. Bacterial ribosomes begin translating an mRNA before RNA polymerase has finished transcribing it. Multiple ribosomes attached to an mRNA form a polyribosome.
RNA polymerase is used to translate DNA to – mRNA
What synthesizes mRNA? –
In eukaryotes, a group of proteins called transcription factors (include many proteins) bind to the DNA promoter, thus initiating transcription –
Many of the eukaryotic promoters include a unique sequence called the TATA box, centered about 30 base pairs upstream of the transcription start site.
The transcription of eukaryotic genes by RNA polymerase generates a primary RNA transcript that contains exons and introns.
The TATA box , (also called Goldberg-Hogness box) –
a DNA sequence (cis-regulatory element) found in the promoter region of genes in archaea and eukaryotes; approximately 24% of human genes contain a TATA box within the core promoter.
Considered to be the core promoter sequence, it is the binding site of either general transcription factors or histones (the binding of a transcription factor blocks the binding of a histone and vice versa) and is involved in the process of transcription by RNA polymerase.
An intron is –
any nucleotide sequence within a gene that is removed by RNA splicing while the final mature RNA product of a gene is being generated. The term intron refers to both the DNA sequence within a gene and the corresponding sequence in RNA transcripts
Sequences that are joined together in the final mature RNA after RNA splicing
Splicing only occurs in –
eukaryotes (The cutting of the introns is caslled splicing)
Genes can have several / many exons –
This is why, we all have the same GENOME but it is translated into different proteins, and so it does different jobs for us. In the process of a cut, INTRONS are taken out, and it is reorganized.
alternative splicing –
A single gene can contain numerous exons and introns, and the exons can be spliced together in different ways. For example, if a gene contains 10 exons, one version of the mRNA transcribed from that gene might contain exons 1-9. Another version of the mRNA might contain exons 1-8, and exon 10.
alternative splicing can produce –
different forms of a protein from the same gene. The different forms of the mRNA are called transcript variants, splice variants, or isoforms.
Caps and Tails for mRNA in Eukaryotes –
Primary RNA transcripts are also processed by the addition of a 5′ cap and a poly(A) tail. With the addition of cap and tail and completion of splicing, processing of the primary RNA transcript is complete.
The product is a mature mRNA. The 5′ cap serves as a recognition signal for the translation machinery. The poly(A) tail extends the life of an mRNA by protecting it from degradation
the binding site for RNA polymerase –
RNA polymerase locates ______ of genes in DNA –
RNA polymerase moves along the template strand in –
3′ to 5′ direction
RNA polymerase synthesizes RNA product in –
5′ to 3′ direction
RNA polymerase does not proofread its work –
The first base transcribed as RNA is defined as the +1 base of that gene.
•To the left ( 5′, or upstream) of this starting point for transcription, -1, -2, …
•To the right ( 3′ or downstream) of this starting point, +1, +2, …
•When RNA polymerase reaches a terminal signal Transcription stops.
When a base is changed, you get a POINT MUTATION, which causes the others to move down one…so the whole three bases sequence gets shifted to another frame, and they become different bases
Genetic codes determine the sequence of amino acids –
One codon only codes one amino acid;
•It is nearly universal, almost all codons specify the same amino acids in all organisms REMEMBER THE BIG CODON CHART
•when several codons specify the same amino acid, the first two bases are <<<<>>
Genetic code – group of three bases that specifies a particular amino acid
64 codons coded for each of the 20 amino acids – (It is redundant. All amino acids except two are encoded by more than one codon).
signifies the start of the protein-encoding sequence in mRNA –
start codon (AUG)
Three stop codons (UGA, UAA, and UAG) in the genetic code –
signal the end of the protein-coding sequence
Translation results in –
1. initiation; initiation factors (IF)
2. elongation; elongation factors (EF)
3. termination. releasing factor
Peptide bond formation –
Newly synthesized amino acid wll bind together through a PEPTIDE BOND (covalent)
The termination phase starts when the A site encounters a stop codon –
This causes a protein called a release factor to enter the site. Release factors resemble tRNAs in size and shape but do not carry an amino acid. These factors catalyze hydrolysis of the bond linking the tRNA in the P site with the polypeptide chain.
A cell does not express all of its genes all of the time –
Instead, they are very selective about the genes they express, how strongly they are expressed, and when they are expressed.
Gene expression occurs when a gene product is actively being synthesized and used in a cell –
Regulation of gene expression is critical to the efficient use of resources and thus survival
DNA – GENE –
( 300 to 1 mil. a segment of DNA – like a recipe ) – Proteins (e.g., hemoglobin)
Suppose that you have a reasonable candidate locus for the disease you are studying. Which of the following would be least likely to contribute useful information? –
b.Single strand conformation polymorphism (SSCP) analysis
d.Denaturing gradient gel electrophoresis (DGGE)
e.Chromosome karyotype e.Chromosome karyotype
Gene Regulation in Eukaryotes is
VERY TIGHTLY CONTROLLED –
Like prokaryotes, eukaryotes can control gene expression at the levels of transcription, translation, and post-translation
Three additional levels of control are unique to eukaryotes –
3.Regulation of mRNA life span or stability.
CHROMATIN remodeling (negative charge) –
The acetyl group causes loosening of the chromatin – repels the histone
The METHYL group makes the gene INACTIVE
Two major types of protein are involved in modifying chromatin structure –
1.ATP-dependent chromatin-remodeling complexes reshape chromatin (They contact DNA on the surface of nucleosome, and twist DNA away from the histone protein core, so that it can be transcribed).
2.Other enzymes catalyze the acetylation (addition of ACETYL groups) and METHYLATION (addition of methyl groups) of histones. (Acetylation of histones is usually associated with activation of genes. Methylation can be correlated with either activation or inactivation).
Regulation of mRNA life span or stability –
Not all mRNA move into the cytoplasm to be translated. There is another way to control them. THEY GET CHOPPED UP. WILL NOT BECOME A PROTEIN.
miRNA kills them
MICRO RNA (miRNA)
In the cytoplasm, additional regulatory mechanisms control gene expression. The stability of mRNAs in the cytoplasm is highly variable. Some are degraded rapidly, allowing for only a short period of translation, while others are quite stable. In many cases, the life span of an mRNA is controlled by RNA interference
•Eukaryotic DNA is packaged with proteins into structures that must be opened before transcription can occur. – •In eukaryotes, transcription is triggered by regulatory proteins that bind to the promoter and to sequences close to and far from the promoter.
•Once transcription is complete, gene expression is controlled by: Alternative splicing; Regulation of the life span of mRNAs.
•Cancer can develop when mutations disable genes that regulate cell-cycle control genes
Which of the following are removed from mRNAs during processing?
3. RNA cap structure
4. Poly (A) tail –
Transcriptionally inactive genes –
1. Are always located within euchromatin
2. Are not located within nucleosomes
3. Are often methylated
4. Are not resistant to DNase 1
(Methyl group makes chromatin more condensed) – 3 Remember what makes them loose? Acetyl groups!
Which codon serves as the start codon in mRNA for translation?
4. UGG –
what phase of cell cycle is DNA copied? –
Nucleic acids are polymerized nucleotides –
A condensation reaction forms a phosphodiester linkage (phosphodiester bond) between the phosphate group on the 5′ carbon of one nucleotide and the -OH group on the 3′ carbon of another.
Types of nucleotides involved: –
1.Ribonucleotides, which contain the sugar ribose and form RNA
2.Deoxyribonucleotides, which contain the sugar deoxyribose and form DNA
T/F RNA and DNA have an OH on 3′ carbon.
Compare DNA synthesis with RNA synthesis – Direction of synthesis: 5′ to 3′ for both
Template: DNA for both
Substrates: DNA: dATP, dGTP, dCTP, dTTP (Deoxy-thymine triphosphate T), RNA: ATP, GTP, CTP, UTP (Uracil – Uridine-5′-triphosphate)
Primer: DNA YES, RNA no
Proof reading: DNA- YES, RNA-no
DNA – dGTP Deoxyguanosine triphosphate is GUANINE
dATP Deoxyadenosine triphosphate is ADENOSINE
dCTP Deoxycytidine triphosphate is CYTOSINE
dTTP Deoxythymidine triphosphate is Thymine
RNA – GTP Guanosine triphosphate is GUANINE
ATP Adenosine triphosphate is ADENOSINE
CTP Cytidine triphosphate is CYTOSINE
UTP Thymidine triphosphate is Thymine
Which of the following leads to a point mutation?
1. Deamination of a cysteme base into a uracil base (c – u)
2. Benzo (a) pyrene conversion of guanine to a thymine base (c – t)
3. Deamination of a 5-methyl cystene into thymine (c -t)
4. All of the above –
When p53 is activated, during severe DNA damage, which of the following occurs?
1. It induces apoptosis
2. It is a transcription factor
3. It serves as a tumour suppressor
4. All of the above –
Transcription in bacteria –
Initiation is the first phase of transcription. However, RNA polymerase cannot initiate transcription on its own. Sigma, a protein subunit, must first bind to the polymerase. Sigma and RNA polymerase together form a holoenzyme, an enzyme made up of a core enzyme and other required proteins.
Sigma acts as a regulatory factor, guiding RNA polymerase to specific promoter sequences on the DNA template strand. All bacterial promoters have a -10 box and a -35 box from the transcription start site (the +1 site) , the remainder of the promoter sequence varies.
RNA Processing in Eukaryotes – •In bacteria, the information in DNA is converted to mRNA directly.
In eukaryotes, however, the product of transcription is an immature primary transcript, or pre-mRNA. Before primary transcripts can be translated, they have to be processed in a complex series of steps.
Eukaryotes – The poly A tail comes AFTER the Stop codon
mRNA carries codons to synthesize proteins, what else can you find inside a MATURE mRNA?
cap and tail and U.T.R.
5′ in front -upstream of of the start codon.
Following the stop codon is 3′ UTR
Following the 3′ UTR is a Poly (A) tail
Key concepts – After RNA polymerase binds DNA with the help of other proteins, it catalyzes the production of an RNA molecule whose base sequence is complementary to the base sequence of the DNA template strand.
Eukaryotic genes contain regions called exons and regions called introns; during RNA processing, the regions coded by introns are removed, and the ends of the RNA receive a cap and tail
A glycoprotein that helps animal cells attach to the extracellular matrix.
at the 3′ end of each tRNA is the binding site for amino acids.
The triplet on the loop at the opposite end is the anticodon that base pairs with the mRNA codon.
A tRNA covalently linked to its corresponding amino acid is called an aminoacyl tRNA.
Gene Regulation in Prokaryote Transcriptional control occurs when the cell does not produce mRNA for specific enzymes. (slow but efficient) MOST COMMON
Translational control allows the cell to prevent the translation of an mRNA molecule that has already been transcribed. (quickly change which proteins are produced)
Post-translational control occurs when the cell fails to activate a manufactured protein by chemical modification. (most rapid response but is energetically expensive).
RNA processing –
1. Alternative Splicing of mRNAs (leads to production of different proteins) 2. Adding Caps and Tails to mRNA
Abnormal gene expression –
Abnormal regulation of gene expression can lead to developmental abnormalities and cancer. Many cancers are associated with mutations in regulatory transcription factors. Tumor suppressor genes slow or stop the cell cycle. If a mutation disrupts the normal function of a tumor suppressor gene, cells are released from this negative control of the cell cycle.
Cellular protein synthesis proceeds in which direction? –
1. Carboxyl to amino terminus
2. Amino to carboxyl terminus
3. 3′ to 5′
4. 5′ to 3′ –
Most eukaryotic genes are controlled at the level of
1. Transcription initiation
2. Transcription elongation
3. Transcription termination
4. Translation initiation –
On what end are the DNA bases added
Why is it called the lagging strand?
What does RNA primase do on the lagging strand?
What is the problem that occurs? –
3′ end ALWAYS
Because it lags behind the other side
It adds an RNA first, so that DNA polymerase can continue
It cant connect it becuase DNA bumps into RNA..so you need DNA ligase to fix them together..it cleans up the messy juctions and lays DNA inbetween
Provides a starting point of RNA (or DNA) for DNA polymerase to begin synthesis of the new DNA strand.
DNA polymerase requires a primer—which consists of a few nucleotides bonded to the template.
leading strand, or continuous strand, -leads into the replication fork and is synthesized continuously in the 5′ 3′ direction – The other DNA strand is called the lagging strand because it is synthesized discontinuously in the direction away from the replication fork and so lags behind the fork.
A Punnett square is now used to –
predict the genotypes and phenotypes of the offspring from a cross.
Autosomal dominant inheritance –
Males and females are affected in roughly equal frequencies; Diseases are observed in Multiple generations
Autosomal recessive inheritance –
An autosomal recessive disorder means two copies of an abnormal gene must be present in order for the disease or trait to develop;
•Are seen only in one generation of a pedigree
•Males and females are roughly equally affected.
•An affected offspring resulted from a consanguineous mating
Sex Linkage Inheritance –
Y chromosomes pair with the large X chromosome during meiosis I. Sex chromosomes pair during meiosis I and then segregate during meiosis II. This results in gametes with either an X or a Y chromosome. Females produce all X gametes.
X-linked recessive inheritance –
If males are much more likely to have the trait, it is usually X-linked.
•X-linked genes are never passed from father to son
•Hemophilia is an example of an X-linked trait resulting from a recessive allele.
-These traits usually skip generations in a pedigree.
•X-linked dominant traits rarely skip generations.
-These traits are indicated in a pedigree wherein an affected male has all affected daughters but no affected sons.
– strange question!!
Phenotype is seen only in males
e. Turner syndrome –
•Suppose you are attempting to find a disease-causing gene, and you have identified a number of families in which the disease is transmitted. If you have no knowledge of the gene product and no reasonable candidate locus, which of the following would be the first technique you would be most likely to use? –
c.Single strand conformation polymorphism (SSCP) analysis
d.Denaturing gradient gel electrophoresis (DGGE)
e.Fluorescence in situ hybridization (FISH) –
A 30-year-old woman has produced two children with down syndrome, and she has also had two miscarriages, which of the following would be the best explanation? –
a.Her first cousin has Down syndrome
b.Her husband is 60 years old
c.She carries a reciprocal translocation involving chromosomes 14 and 18
d.She carries a Robertsonian translocation involving chromosomes 14 and 21
e.She was exposed to multiple X-rays as a child –
which one of the following is a sex chromosome aneuploidy
d.Patau Syndrome. –
The severe form of alpha 1 antitrypsin deficiency is the result of a single nucleotide substitution that produces a single amino acid substitution. What is best description of this?
e.Splice-site mutation –
When DNA replication starts..what breaks? –
Hydrogen bonds between the nucleotides of two strands
The elongation of the leading strand during DNA synthesis depends on –
the action of DNA ploymerase
True replication of DNA is due to –
complementary base pairing rule
A crossover event that places the SRY gene on the X chromosome can cause
a.an XX male
b.a male who has features similar to those of Klinefelter syndrome
c.an XY female
d.all of the above –
DNA synthesis can be specifically measured by estimating the incorporation of radio labelled –
During mitosis, ER and nucleolus begin to disappear at what stage? –
Which is the best stage to view the shape, size and number of chromosmes? –
The number of chromatids at metaphase is –
two each in mitosis and meiosis
During which phase of mitosis would you expect to see chromosomes consisting of two chromatids within the nucleus? –
prophase P-MAT Sister chromatids form during interphase and remain as the nuclear envelop breaks down during prophase.
Mode of DNA replication in E.coli is –
semiconservative and bidirectional
RNA primary structure –
Like DNA, RNA has a primary structure consisting of a sugar-phosphate backbone formed by phosphodiester linkages and, extending from that backbone, a sequence of four types of nitrogenous bases. The primary structure of RNA differs from DNA in two ways:
1.RNA contains uracil instead of thymine.
2.RNA contains ribose instead of deoxyribose. The presence of the -OH group on ribose makes RNA much more reactive and less stable than DNA.
Is there a primer for RNA transcription (mRNA)? –
No. The ENZYME that performs transcription is referred to as a DNA-dependent RNA Polymerase. No primer.
Q. What is the direction of newly synthesized DNA? –
All is 5′ to 3′
But in terms of the bubble, the DIRECTION is different. From righ to left. Both 5′ to 3′ and 3′ to 5′ can be templates.
DNA Polymerase 2 –
Builds a new duplex DNA strand by adding nucleotides in the 5′ to 3′ direction. Also performs proof-reading and error correction.
Mitosis is similar to –
Using linkage analysis, you have mapped a disease gene to a 1 megabase (Mb) region of a specific chromosome. Which of the following approaches would be least useful in identifying and cloning the gene?
a.Single strand conformation polymorphisms (SSCP) analysis
d.Radioactive in situ hybridization –
In assessing a patient with osteogenesis imperfecta, a history of bone
fractures, as well as blue sclerae are noted, these findings are an example of
Eukaryotes differ from prokaryotes in mechanism of DNA replication due to –
discontinuoys rather than semidiscontinuous replication
What are RNA Polymerase 1, 2, 3? –
they synthesize rRNA, mRNA, and tRNA respectively
Mitotic anaphase differs from metaphase in possesing –
same number of chromosomes as and half number of chromatids
What is a specific segment of DNA that directs protein synthesis? –
DNA has two different sized grooves: –
the major groove and the minor groove.
Sometimes, information flows in the opposite direction. From RNA, back to DNA. –
Indeed, evidence has recently emerged that at least some RNA-mediated epigenetic changes can be inherited. In this way, environmental influences could, via RNA editing, influence the phenotype of the next generation, at least epigenetically. This would be – an echo of the ‘inheritance of acquired characteristics’ model proposed by Lamarck more than 200 years ago
DNA replication –
is the process of producing two identical copies from one original DNA molecule. This biological process occurs in all living organisms and is the basis for biological inheritance. DNA is composed of two strands and each strand of the original DNA molecule serves as template for the production of the complementary strand, a process referred to as semiconservative replication. Cellular proofreading and error-checking mechanisms ensure near perfect fidelity for DNA replication.
Directionality has consequences in DNA synthesis, because DNA polymerase can synthesize DNA in – only one direction by adding nucleotides to the 3′ end of a DNA strand
Which of the following is a functional element of a plasmid?
a. origin of replication
b. drug-resistance gene
c. polylinker sequence
d. a and b
e. all of the above –
DNA Secondary Structure –
James Watson and Francis Crick determined:
1.DNA strands run in an antiparallel configuration.
2.DNA strands form a double helix. The hydrophilic sugar-phosphate backbone faces the exterior. Nitrogenous base pairs face the interior.
DNA Secondary Structure 5 – 3.Purines always pair with pyrimidines. Specifically, strands form complementary base pairs A-T and G-C. A-T have two hydrogen bonds. C-G have three hydrogen bonds.
4.DNA has two different sized grooves: the major groove and the minor groove.
Nucleosomes (small) and Chromatin (BIG)
Histone spools –
Nucleosome home –
Chromatin city –
What does the BEADS on a STRING idea mean?
STRING – Chromatin the big city
BEADS – Nucelosomes the homes in the city – STEPS
1. Histones: Eight separate HISTONE subunits (big protein sticky balls) attach to the DNA molecules..they are like a spool for the DNA to wind around.
1. Nucleosome: The combined TIGHT LOOP of DNA and PROTEIN
2. Chromatin: Multiple nucleosomes, stacked on top of each other (30nm thick)
3. Chromosomes: looped again and further packaged (6 feet of DNA fit into each cell of our body)
Chromosomes form only when cells are dividing. At the end of cell division, it becomes less higly organized.
e.g., Interphase, metaphase
Chromatin city is the combination of DNA and histone proteins; DNA wraps around histone proteins spools
-the “beads on a string” structure. 10nm chromatin
-30 fiber form loops attached to scaffolding proteins
-Higher order packaging-Heterochromatin
Central Dogma Key concepts: –
1.Nuclei acids (DNA and RNA);
2.Nucleotides (nucleoside monophosphates) linked by phosphodiester bonds;
3.Directional (sequence always 5′ to 3′);
4.Double stranded DNA ( associated by hydrogen bonding, sequences are complementary to each other);
5.In eukaryotic cells, DNAs are in nucleus;
6.nucleosome-DNA packed with histones-euchromatin;
7.Highly packed chromatin forms heterochromatin (no gene expression);
8.Difference between DNA
9. Central dogma
DNA Helicase –
Also known as helix destabilizing enzyme. Unwinds the DNA double helix at the Replication Fork.
Semiconservative replication – would produce two copies that each contained one of the original strands and one new strand.
Why is it called a phosphoDIester bond? –
Because there are two bonds. One is the bond between the phosphate group on the 5′ carbon of one nucleotide (look at it..you can see that the phosphate is attached to the 5th carbon of the sugar- don’t be impressed, it was already attached to the nucleotide). And the 3′ carbon on another sugar (look to see how the top of the phosphate is linked to the bottom part of the sugar above it.)
Notice that the hydroxyl group (OH) of the phosphate is reacting with the Hydroxyl group (OH) on the 3′ to make water…so two H and one O make water.
The second is the bond with the sugar phosphate backbone. On the OTHER side!!
Notice that the 5′ end has an unlinked 3′ carbon.
And the other end has an unlinked 3′ carbon. – Why is it always a 5′ to 3′ building pattern? B/c the 3′ direction would not allow anything to build on it. There is a CH2 group in the fig below.
DNA’s secondary structure consists of two antiparallel strands twisted into a double helix – DNA can store and transmit biological information. The language of nucleic acids is contained in the sequence of the bases.
DNA is water soluble. It carries a negative charge. (phosphate carries a negative charge)
DNA plus Histone – NECLEOSOME
The Central Dogma of Molecular Biology
Where is mRNA made? –
In the nucleus.
Protein synthesis takes place in the cytoplasm.
DNA (information storage) Transcription mRNA (information carrier) Translation Proteins (active cell machinery)
Replication is Semi-conservative and bidirectional; – multiple origins of replication in eukaryotes have multiple replication bubbles. A replication fork is the Y-shaped region where the DNA is split into two separate strands.
Topoisomerase def. –
Relaxes the DNA from its super-coiled nature.
Is there a primer for DNA replication? –
In order to undergo DNA replication –
the double strand must split into a single strand…which becomes a template.
DNA replication happens in the nucleus
What is the template during transcription?
What is responsible for the transcription? –
What is the difference between RNA and DNA polymerase?
No primer for RNA polymerase
What kind of primer is necessary for DNA replication?
What is the direction / orientation of newly synthesized mRNA?
5′ to 3′
DNA replication –
In a cell, DNA replication begins at specific locations, or origins of replication, in the genome. Unwinding of DNA at the origin and synthesis of new strands results in replication forks growing bidirectionally from the origin. A number of proteins are associated with the replication fork which assist in the initiation and continuation of DNA synthesis. Most prominently, DNA polymerase synthesizes the new DNA by adding complementary nucleotides to the template strand.
•A 16-year-old girl is only 1.5 meters tall, with no breast development, and no menstruating. She is doing OK at school. What is the most likely underlying basis for her condition?
a.45, X karyotype
b.A balanced reciprocal translocation
c.A balanced Robertsonian translocation
d.Two Barr bodies
e.Deletion of an imprinted locus –
For DNA replication, can both strands be the template? –
Can both strands be used as a template for transcription? –
No, only one strand
Where is the stop codon?
On the DNA template strand.
What is another word for translation? –
Where does it happen?
What synthesizes mRNA?
The first step in converting genetic information into proteins is transcription, the synthesis of an mRNA version of the instructions stored in DNA. –
RNA polymerase performs this synthesis by transcribing only one strand of DNA, called the template strand.
The other DNA strand is called the non-template, or coding strand, which matches the sequence of the mRNA, except that RNA has uracil (U) in place of thymine (T).
RNA polymerase performs a template-directed synthesis in the 5′ to 3′ direction. But unlike DNA polymerases,
RNA polymerases do not require a primer to begin transcription.
Transcription and Translation in Eukaryotes –
In eukaryotes, transcription and translation are separated. mRNAs are synthesized and processed in the nucleus and then transported to the cytoplasm for translation by ribosomes.
What do restriction enzymes do –
CUT DNA, and replice with H bonds glue
Incomplete penetrance –
Some individuals who have the disease genotype, do not have the disease phenotype
Mitochondrial inheritance •Diseases are usually affect nerve and/or muscle.
•Mitochondrial DNA is inherited exclusively through females
•All offspring of an affected females are affected.
•Non of the children of an affected male is affected.
Character of single-gene disease def –
•Symptom severity varies
•Incomplete penetrance –
some individuals with the disease-causing mutation who do not exhibit clinical symptoms.
•Locus heterogeneity –
one disease caused by mutations in different loci
•Random new mutation
the symptom is more severe in new generation than in older generation.
only maternal allele active for one locus, and only paternal allele active for another locus. (eg Prader-willi and Angelman Syndromes the gene from one parent is inactivated due to normal imprining, the gene from the other parent is mutated).
1.Basic terms and new methods in cytogenetics.
2.Numerical chromosome abnormalities;
3.Structural chromosome abnormalities;
Chromosome nomenclature –
p=short arm of the chromosome
q=Long arm of the chromosome
t=translocation; del=deletion; + extra – Missing X, Y sexchromosome 1-22 aotuosome
Thick arrow-one spot-deletion of 17p – smith-magenis syndrome –
Cytogenetics (a branch of genetics)
•To study microscopically observable chromosomal abnormalities.
1. using G-banding technique, as well as
2. new molecular cytogenetic techniques
such as fluorescent in situ hybridization (FISH).
In which a chromosome-specific DNA segment is labeled with a fluorescent tag to create a probe. The probe is then hybridized with patient’s chromosomes, which are visualized under a fluorescence microscope. The probe will mark the presence of the chromosome segment being tested.
Human male karyotype (metaphase chromosomes) Karyotype-ordered display of chromosomes
•Routine chromosome analysis .
•Arrest chromosomes at metaphase, and treat with trypsin and followed by Giemsa staining. This creates unique banding patterns on the chromosomes ( G-banding).
Numerical chromosome abnormalities –
•Euploidy: when a cell has a multiple of 23 chromosomes.
1.Haploid: 23 chromosomes, in gametes;
2.Diploid: 23×2=46 chromosomes, in somatic cells;
3.Triploid: 23×3=69 chromosomes, rare, lethal
( 3 copies of each chromosome)
4. Tetraploid: 23×4=92, rare, die after birth
Aneuploidy, gain or loss of a specific chromosome –
•Monosomy (loss of a chromosome)
•Trisomy (gain of a chromosome)
•All autosomal monosomeies die after birth;
•Autosomal trisomies can live: such as
•Trisomy 21 ( 47, XY, +21 or 47, XX, +21) down Syndrome.
•Trisomy 18 ( 47, XY, +18 or 47, XX, +18) Edward Syndrome.
•Trisomy 13 (47, XY, +13 or 47, XX, +13) Patau Syndrome.
Sex chromosome aneuploidy
•Turner Syndrome (45, X or 45, XO) only monosomy can be alive
•Klinefelter syndrome (47, XXY)
What cause aneuploidies? –
•Nondisjunction during meiosis
trisomy 18 (Edwards’ syndrome) –
with typical facial features including a narrow head, short palpebral fissures, and malformed external ears as well as characteristic overlapping of the index finger on top of the middle finger.
full trisomy 13 (Patau syndrome). This baby has a cleft palate, atrial septal defect, inguinal hernia, and postaxial polydactyly of the left hand
A girl with Turner syndrome (45,X). Note the characteristically broad, webbed neck. Stature is reduced, and swelling (lymphedema) is seen in the ankles and wrists.
A male with Klinefelter syndrome (47,XXY). Stature is increased, gynecomastia may be present, and body shape may be somewhat feminine.
Single-gene disorders –
1.Mendel and genes;
7.Characteristic of single-gene disorder
Mendel called the genetic determinant for wrinkled seeds recessive and the
determinant for round seeds dominant. –
In modern genetics, the terms dominant and recessive identify only which
phenotype is observed in individuals carrying two different genetic
Mendel repeated these experiments with each of the other traits. In each case, the
dominant trait was present in a 3:1 ratio over the recessive trait in the F2 generation.
Mode of inheritance: –
Structural chromosome abnormalities –
Translocations: – Translocations:
broken elements of a chromosome reattach to other chromosomes.
Reciprocal translocation-genetic material is exchanged between non-homologous chromosomes. Most balanced translocation carriers are healthy. However, carriers of balanced reciprocal translocations have increased risks of creating gametes with unbalanced chromosome translocations leading to miscarriages or children with abnormalities.
Both DNA and RNA have an OH group on which carbon?