Lecture Notes for Biology 202-Dr. McGuire's portion of the class

Text:  Life, the science of biology by Purves et al., 6th edition

CHAPTER 9:  Meiosis

The cell cycle:  figure 9.4 review
The packing of the chromatin figure 9.7--2 meters of DNA packed into 5µm nucleus; even more packing occurs in chromosome
Mitosis:  requires duplication of DNA.  Sister chromatids and centromers for mitotic center (organizes microtubules in center); cytokenesis-part of M phase at end of nuclear division  figure 9.8
Purpose of mitosis--to maintain genetic material

Meiosis:  sexual reproduction; results in diversity
2 parents each providing gamete with 1/2 DNA of organism
2 gametes make a zygote or fertilized egg

Chromosomes come in homologous pairs:
Haploid cells- one copy of each chromosome = 1n
Diploid cells- two copies of each chromosome = 2n
Humans have 46 chromosomes; 22 autosomes (2 of each) and 1 sex chromosome (2 of each).  The 2 sex chromosomes are homologous or not depending.
Karyotype = number, shape, and sizes of chromosomes in a single cell.

Meiosis:  figure 9.14
Prophase = identical to mitosis BUT homologous chromosomes pair to form a synapse.  The four sister chromatids make a structure called the tetrad or bivalent.  This is unique to meiosis and does not happen in mitosis.  This is a key event that makes meiosis different.
2 rounds of division in meiosis--the other thing that separates it from mitosis.  First meiotic division reduces the # of chromosomes to 23 (but each chromosome consists of 2 sister chromatids).  Second meiotic division separates the sister chromatids and reduces DNA content to 1n.  Meiosis 2 looks like mitosis except:  no further DNA replication occurs, crossing over (below) can occur, and humans have 23 chromosomes on the equatorial plate at this phase instead of the 46 present in mitosis.

In meiosis one, the alignment of the homologous pairs is random; some maternal and some paternal chromosomes go to each side of the plate, resulting in random assortment.

In homologous pairs, attachments can form.  Attachments between sister chromatids from different members of the homologous pair--form chiasmata--these are sites of crossing over or homologous recombination.  Figure 9.16

Mitosis versus meiosis review--figure 9.17

Figure 9.18= meiotic errors.  Nondisjunction leads to aneuploidy (trisomy etc.).  Most common in humans is trisomy 21 (Downs).  Trisomy 13, 15, and 18 also viable but die early.  Translocations result in partial trisomys.  Some viable as well.  Other species are not as sensitive as us to nondisjunction.

Cell Death:  Table 9.2; figure 9-19.

What happens to cells that are genetically defective, either after mitosis or meiosis, or in some other way compromised?
Answer--cell death
2 ways for cell to die--necrosis and apoptosis

Necrosis is very damaging, induces inflammation which causes more cell death, it is not controlled and it is found in disease situations.

Apoptosis, on the other hand, is genetically programmed (known as programmed cell death), controlled, does not induce inflammation and is an active process.  The cell is phagocytosed before lysis; no injury is caused to surrounding cells.  It is recognized as apoptotic by macrophages.

NOTE:  table 9.2 is wrong.  In necrosis, phagocytosis is of debris, not of cells.  Phagocytosis of apoptotic cell occurs before debris generated, by surrounding MACROPHAGES.
 

CHAPTER 10:  Mendel

Mendel bred "true-breeding" plants in figure 10.1.  to get F1 generation.  Wanted to determine which traits were dominant and which were recessive.
Can breed F1 plants to get F2 generation--thus study traits through multiple generations.

Table 10.1--some of the traits analyzed by Mendel.
Monohybrid crosses-figure 10.3.  When breed F1 back to itself, recessive traits reappear at 3:1 ratio.

Table 10.1--all monohybrid crosses showed 3:1 ratio of Dom:rec at all traits.

These traits--we call genes--different forms of genes are called alleles.
Homozygous--true-breeding due to fact they only have one allele.
Heterozygous--F1
Figure 10.4--phenotype versus genotype
Figure 10.5--follow chromosomes through meiosis to understand segregation of traits.
Figure 10.6-- the back cross or "test-cross".  Answers the question, is a plant homozygous or heterozygous?

Independent assortment:  studied by dihybrid crosses and shown in figures 10.7 for traits and 10.8 for chromosomes.  Stress the 9:3:3:1 ratio of phenotypes in this cross.

Inheritance follows the rules of probability:  figure 10.9.
Go over a pedigree describing inheritance of dominant allele (figure 10.10) versus one for recessive allele in figure 10.11.

Variations on the theme:
Incomplete dominance:  phenotype ratio same as genotype ratio.  Gene dosage important here.  figure 10.13
Co-dominance--use ABO blood group as example, figure 10.14
Pleiotropic:  1 gene responsible for two different traits (coat-color and crossed eyes in Siamese cats)
Epistasis:  the phenotype of one gene is influenced by another gene (figure 10.17).
Hybrid vigor:  based on minimizing effect of detrimental alleles.
Penetrance:  proportion of individuals with an allele that show the expected phenotype.

Linked genes:  Figures 10.19-22.  Work the problem in Figure 10.24 with the students and tell them to practice question in study guide.

Sex-linked genes:  figures 10.25 and 10.26.
Not every species uses the XY system.  Humans--Y denotes male.  In Drosophila, 2 X denote female even if Y present.  This is a dosage effect.  Birds:  female has 2 different chromosomes (called ZW) while male is ZZ.  Some insects 2X = female, 1X = male.
New rule here:  in XY system, male is always hemizygous for X and Y encoded genes.
 

CHAPTER 11  DNA

Figures 11.1-3:  Genetic material can be transferred from one bacteria to another, but was it DNA or protein that made up genetic material?  Used bacteriophage to answer question.  Answer:  DNA

Figure 11.5:  base pairing; 1 purine/1 pyrimidine in each base pair.
3 dimensional structure of DNA:   figure 11.6-7.  (Picture a ladder in your mind and twist it into a helix)
Strands are anti-parallel, form a minor groove (the space created by the "rungs" of the ladder) and a major groove.
Draw a base and remind them of the 5' and 3' ends of the base.  Stress 5' to 3' directionality to DNA and the anti-parallel structure.

DNA as genetic material has 4 functions:  fulfilled by double-stranded structure.
1)  stores genetic information
2)  must be able to change at a low frequency
3)  must replicate precisely most of the time
4)  must be expressed in the phenotype

How does DNA replicate?  Answer:  semi conservatively; figures 11. 8-10

Figure 11.11:  DNA is synthesized 5' to 3' (grows from the 3' end when 5' PO4 of incoming nucleotide attacks 3' OH of growing DNA strand.
To make DNA:  need template, a growing strand (called a primer), free nucleotides, and DNA polymerase (the enzyme that synthesizes DNA)
DNA polymerase has 3 requirements:  3' to 5' template strand, a primer, and free nucleotides.  Synthesizes DNA in the 5' to 3' direction (all polymerases work in this direction)

Origins of Replication:  figure 11:12-13   DNA move through the origin, the origin does not move down DNA.  DNA synthesis goes in both directions from the origin; 2 replication forks per origin; both strands serve as template.  (FYI;  1 chromosome = 80 million base pairs).  Figure 11.15

Replication complex has 5 components:
DNA helicase:  opens the double helix
Single strand (ss) DNA binding proteins:  stabilize the single stranded DNA
RNA primase:  creates the RNA primer used by DNA polymerase
DNA polymerase:  adds nucleotides complementary to the template strand, checks (proof-reads) and repairs if it screws up.
DNA ligase:  seals up breaks in the sugar-phosphate backbone.

Figure 11:16-17:
1.  DNA helicase unwinds DNA-opens the two strands, uses ATP for energy
2.  ss DNA binding proteins keep the strands apart.
3.  DNA polymerase needs to add dNTPs to a growing strand; therefore, RNA primase makes RNA primer using template DNA strand.
4.  DNA polymerase (pol III in bacteria) adds dNTPs to RNA primer in a 5' to 3' direction.  Makes the 5' to 3' strand using the 3' to 5' template as one continuous strand--called leading strand synthesis.  On the 3' to 5' template, however, synthesis also has to go 5' to 3' and the template is only available for so far due to the double stranded nature of the DNA.  So RNA primase has to keep binding and making another primer every time the template is unwound.  At each primer, DNA polymerase binds and synthesizes the DNA until it runs into a previous primer.  These fragments of RNA primer with DNA are called Okasaki fragments and this strand is called the lagging strand since its synthesis lags behind the other strand.
5.  figure 11.18.  The primers are removed by DNA polymerase (by pol I in bacteria) and DNA polymerase uses the upstream Okasaki fragment to extend the DNA through the sequence previously occupied by the primer.
6.  DNA ligase seals the nick in the sugar phosphate backbone to seal the two fragments of DNA together.
Steps 5 and 6 also occur on the leading strand at each origin on a eukaryotic chromosome which has hundred of origins.

Proofreading and repair:  figure 11.19
DNA polymerase, helicase, ligase and other proteins recognize mistakes.  Proofreading lowers the number of mistakes from 1 in 1 million to 1 in 1 billion.
Two types of repair if mistake is not caught during synthesis or if it occurs from DNA damage:   mismatch and excision repair.
 

CHAPTER 12:  From gene to protein

The original concept:  one gene; one enzyme.  Figure 12.1
The central dogma of molecular biology:  figure 12.2; exceptions in retroviruses, RNA viruses and prions.

Figure 12.3:  overview of gene expression to protein production.

Transcription:  figure 12.4.  Formation of a RNA molecule from DNA, using RNA polymerase.  Requires 3' to 5' template strand and nucleotides.  Same process makes mRNAs, tRNAs, and rRNAs but there are three different RNA polymerases.
Initiation:  begins at a promoter-seq of DNA that RNA pol binds to.  Directs which strand will be copied (always 3' to 5'); contains RNA initiation site where transcription begins; need other proteins for Pol II docking (chapter 14).
Elongation:  RNA pol II unwinds the DNA ~ 20 bp at a time; adds new nucleotides to the 3' end of the growing strand.  RNA transcript is anti-parallel (complementary) to the template.  No proof-reading in Pol II--make a mistake 1 in every 10,000 to 100,000 base pairs.  Why isn't this a problem?
Termination:  RNA pol stops.  RNA is processed in eukaryotes in nucleus before being transported to the cytoplasm where it will be translated.

The Universal genetic code:  figure 12.5--show them how to read.  TABLE USES SEQUENCE IN mRNA

Translation:  uses the universal genetic code and is the production of proteins from a mRNA.
Uses essentially 3 letter non-overlapping "words" called codons.
AUG is the start codon and also encodes the amino acid methionine.  Every protein we make starts with methionine.
UAA, UAG, and UGA--all stop codons; causes ribosome to stop translating.
Code is redundant--several codons can encode the same amino acid.
Code is NOT ambiguous:  single codon always encodes the same amino acid.  (figure 12.6)
Genetic code is complementary to the coding strand (the 3' to 5' strand); the mRNA is identical in sequence to the non-coding strand of DNA but it contains U not T residues.

Preparation:  figures 12.7-8
Translation by ribosomes requires tRNAs that carry amino acids be present in cell.
tRNAs have an "anti-codon" sequence and the anti-codon determines the amino acid that will be carried by the tRNA.
tRNA stands for transfer RNAs and link the codon in the RNA to specific amino acids using the code.
Wobble is the observation that the 3rd base of the codon can see tRNAs with different bases in the 5' position.  So the 3rd base is less important in determining the amino acid in many but not all codons.  For instance, GCA, GCC, GCU are all alanine codons that interact with 1 tRNA

Figure 12.8:  the amino acyl tRNA synthetase:  If you understand these enzymes, you can understand the universal genetic code.  Binds a specific amino acid and ATP; the amino acid gets phosphorylated and the amino acid is now considered "activated"; then a specific tRNA for that amino acid binds to the tRNA binding site (specificity determined by the anti-codon) and the amino acid is linked to the tRNA at the amino acid binding site (CCA) on the 3' end.  Now the tRNA is called a charged tRNA.  Anticodon is considered the second genetic code.

Ribosome:  figure 12.9  large subunit made up of 3 rRNAs and 45 different proteins while small subunit is made of 1 rRNA and 33 different proteins.
4 tRNA binding sites:  the T site binds charged tRNAs bringing in amino acid encoded by next codon; A site--tRNA in T site transfers over and the anti-codon of the tRNA interacts with the codon in the RNA which is situated in the A site.  P site--holds the tRNA with the growing polypeptide chain.  The E site is the exit site where the tRNA in the P site exits when its growing protein is transferred to the tRNA in the A site.

Initiation:  Figure 12.10-small subunit binds to sequences in the RNA upstream (5') of the AUG.  The figure is wrong in that the sequence, known as the ribosome recognition sequence (RRS) or ribosome binding site (RBS), is not the AUG but upstream of the AUG.  The sequence in the mRNA upstream of the AUG are the 5' untranslated region (5'UT) and the sequence downstream of the stop codon in the mRNA is the 3'UT region.  Once the small subunit is bound, the tRNA bearing methionine, with the right anti-codon of course, interacts with the AUG and the large subunit of the ribosome binds to the complex with the met-tRNA in the P site.

Elongation:  Figure 12.11--tRNA needed by the next codon, present in the A site, comes in and interacts with the codon in the A site.  An enzyme activity in the large subunit, called peptidyl transferase, transfers the growing polypeptide chain on the tRNA in the P site to the amino acid on the tRNA in the A site.  Peptide bond is formed such that the protein is now one amino acid longer.  At the same time as the transfer of the protein, the ribosome slides down the mRNA 3 nucleotides, moving the tRNA with the growing protein into the A site and pushing the empty tRNA out through the E site.  This continues to happen as long as the codon in the A site encodes an amino acid.  The protein is growing in the N-terminal to C-terminal direction and always starts with methionine.  Energy for the movement of the ribosome is generated by the hydrolysis of GTP.

Termination:  figure 12.12--when a stop codon is present in the A site, no tRNA will interact.  Instead a protein called release factor binds and causes the ribosome to cleave the protein from the tRNA in the P site.  The protein is released into the cytosol (or ER; see below) and the ribosome detaches from the mRNA.  Translation can begin again.

Neither the tRNAs, the ribosomes or the mRNA are changed in any way by this process.  All of them are recycled by the cell and used again.  tRNAs get new amino acids, mRNAs get translated by other ribosomes and ribosomes translate either the same or a different mRNA.

Polyribosomes:  figure 12.13

Translation occurs either in the cytosol if proteins destined for inside the cell (discuss localization sequences such as nls etc.); or translated on the ER membrane if destined for the cell surface or secretion.  Figure 12.13 and 12.14.  Describe secretory and plasma membrane protein pathways.

 Post-translational events:  figure 12.16

Mutations:  Somatic versus germ-line.
Point mutations versus chromosomal alterations.  Chromosomal alterations shown on figure 12.18.
Point mutations:  synonymous (no change in a.a. sequence), missense (change in a.a. seq.), nonsense (introduction of a stop codon) and frame-shift-addition or deletion of bases in groups not divisible by 3.
Spontaneous versus induced mutations and consequences:  Figure 12.19

EXMAPLE of how to make an mRNA, translate it, and consequences of mutations:

DNA template 3'…..CGGGTACACCGAGGGCCTAATTGGGC….5'

mRNA made   5'…..GCCCAUGUGGCUCCCGGAUUAACCCG….3'

Protein encoded:  met-trp-leu-pro-asp

Synonymous mutation:  3'…..CGGGTACACCGAGGGACTAATTGGGC….5'
Missense mutation:  3'…..CGGGTACACCGAGGGCCAAATTGGGC….5'
Nonsense mutation:  3'…..CGGGTACATCGAGGGCCTAATTGGGC….5'
Frameshift mutation:  3'…..CGGGTACACCTGAGGGCCTAATTGGGC….5'

Have students change mRNA and translate sequences to see the results.  On last one explain the definition of frames and how the last mRNA is "frame-shifted".
 

CHAPTER 13:  Prokaryotic gene expression

Protein content of bacteria changes rapidly depending on the conditions that they find themselves in.  They regulate their gene expression to reflect their growth conditions--saves energy and time and allows bacteria to focus their energy where it is needed.

Respond to "inducers" or inhibitors" in their environment.  Figure 13.13.  Do this using "operons", i.e. groups of genes in related biochemical pathways that are coregulated by "inducer and inhibitors".  Figure 13.14.  Some genes are not in operons and are expressed all the time.  This is called constitutive expression.

Single promoter regulates the production of multiple products all involved in a single biochemical pathway.  End product can negatively regulate either by inhibiting the transcription of the genes (works through proteins called repressors) or can inhibit the activity of an early enzyme in the pathway.

Repressors:  figure 13.15.

Operons:  Always contain 3 components; the promoter (promotes transcription); the operator (binds the regulator or repressor protein); and the structural genes for the enzymes or proteins required for the pathway controlled by the operon.

The Lac operon:  figure 13.16.  Operator always present downstream of the promoter which allows binding of the repressor to block transcription mediated by RNA polymerase.  The purpose of the lac operon is to control the genes required for the use of lactose as an energy source.  When lactose is present, the bacteria need the enzymes to break it down; when lactose is absent they don't.  This controls the operon.
Regulatory gene:  encodes the lac repressor, a molecule that is sensitive to lactose and controls the transcription of the gene.  The regulatory genes can sometimes be found near the operon (in the DNA) but often times are present in another part of the genome.  They are not part of the operon.  The repressors are produced at low levels all the time--they are constitutive.

Figure 13.17.  When lactose is not present, the repressor is in a conformation that is active for DNA binding.  Its binding to the operator blocks RNA polymerase from unwinding the DNA and transcribing the genes because RNA pol binds to the promoter and the repressor is in the way.  When lactose is present, it binds to a lactose binding site in the repressor and changes the conformation.  The repressor can not bind DNA and transcription will occur.

Another example; the Trp Operon, figure 13.18.  This operon is active if the amino acid is missing.  If it is provided by the media, the cell will not need to make trp so the operon will shut off.

Figure 13.19--the control of the lac operon by glucose.  If the cell has both glucose and lactose, it prefers to use glucose because it takes less work to generate energy from glucose than lactose.  So glucose controls the lac operon but the control is indirect.  Low glucose results in the production of cAMP.  cAMP binds to a protein called CRP or cAMP receptor protein.  CRP binds DNA (in the promoter, NOT the operator) when it interacts with cAMP and acts as an enhancer.  An enhancer enhances transcription by recruiting RNA polymerase.  Get more transcription when glucose is low.  When glucose is high, cAMP levels are low.  When cAMP is low, it can't bind to CRP which can't bind DNA and enhance transcription.  Transcription occurs at a very low level but not nearly at the level that occurs with the enhancer.
 

CHAPTER 14; The control of Eukaryotic gene expression.

Table 14.1 and figure 14.1:  Major differences between prokaryotes and eukaryotes.  Eukaryotic genome is much larger and more complex that prokaryotic genome.  The increasing complexity found in organisms is reflected somewhat by the size of the genome but not completely.  Not regulated nearly so nicely as the operon system in prokaryotes.

Figure 14.2:  What we have discussed before but failed to describe the processing events that must occur to a RNA after transcription to make a mature processed mRNA.  These events are linked with transport of the mRNA from the nucleus to the cytoplasm where it can be translated.

Repetitive sequences:  increase in complexity as the complexity of the organism increases.  3 types of highly repetitive DNA=satellites, minisatellites, and microsatellites.  Describe and summarize these (homemade slide).

Telomers:  figure 14.3.  Moderately repetitive.  At the ends of chromosomes in eukaryotic genomes.  One of the things we didn't discuss with the replication of DNA is that the lagging strand of DNA synthesis can not replicate the very end of the chromosome because there is nothing that can serve as a primer.  Remember that primase has to make a RNA primer that DNA pol can use to synthesize new DNA.  Then the primer is removed and the 5' Okasaki fragment is used by DNA polymerase to replace the primer and elongate the DNA strand that is then ligated to the 3' fragment (review in chapter 11).  At the end of the chromosome, when the primer is removed, there is no 5' Okasaki fragment to extend.  Therefore, we lose DNA on each end of the chromosome.  The overhanging bases on the leading strand are deleted to make a flush end.  So in each round of replication, we lose DNA.  This is obviously an intolerable situation because we would start deleting genes after awhile which we can't afford.  Thus the telomer.  In humans, TTAGGG repeated 2500 times at the end of each chromosome.  Loss of telomeres occurs with age; cells eventually lose their ability to divide.  Some cells must divide much more than other cells:  gut epithelium, bone marrow and germline cells (egg and sperm producers).  Express telomerase.  90% of human cancers also express this gene even though most are not supposed to.

Two other moderately repetitive DNAs:  rRNA (figure 14.4) and tRNA genes.  We have 280 copies of the genes for rRNAs.  These genes have no introns and are synthesized all the time.  One last moderately repetitive sequence is the Alu repeat.  There are 300,000 copies of this sequence (digestible by a restriction enzyme called Alu I, hence the name).  These Alu repeats may be the origin of replication.  Obviously the origin of replication, whatever it is, must be a repetitive DNA element since there are hundreds per chromosome.

Transposable elements:  figure 14.5.  Two kinds of transposons; SINES and LINES (short (500 bp) or long (7000 bp) interspersed elements).   SINES are transcribed, not translated.  LINES are transcribed and some are translated.  They constitute 15% of the genome.  Movement is rarely actually observed but some mutations and inherited diseases are caused by transposable elements.  Retrotransposons are retrovirus-like sequences that lack an Env gene so can't make virus.  Rare in humans but they do occur.  Common in animals.  DNA transposable elements:  eg. P element in Drosophila.

Structure of protein coding genes shown in figure 14.6.  Introns=sequences not present in the final mRNA.  Exons=sequences present in the final mRNA.  Yellow denotes 5'UT while red denotes 3'UT.  Both 5' and 3' UTs are encoded by EXON sequences.  Any sequence that ends up in the mature mRNA is exon sequence.  So in figure 14.6 exon 1 includes 5'UT and exon 3 includes 3'UT.  Kind of drawn wrong because exons more than just dark blue sequence.  Introns are sequences that are removed from the precursor RNA in the final mRNA.  Process is called "splicing".

Gene families:  figure 14.9.  Many of our genes are present in multiple copies.  These genes can be encoded in DNA sequences near to each other (in clusters) or dispersed throughout the genome.  Some of the genes in gene clusters or gene families are pseudogenes; they do not encode a functional product (they have a stop codon in the reading frame or some such thing).  Many of the genes encode functional products that work slightly differently from one another.  For example, in the globin gene cluster, we have multiple products that are used at different times in development because we get our oxygen from different places during development (i.e. from the air after birth but from the mother's blood before birth).  Figure 14.10.

RNA processing:  Three types;  capping, tailing and splicing.  This is applicable to RNA pol II transcripts, the RNA polymerase that transcribes our mRNAs.  RNA pol I and III transcripts are NOT processed the same way.

Figure 14.11:  The primary transcript has nothing but the RNA sequence that was copied off the DNA template.
Event 1:  capping.  As the transcript is made by RNA polymerase II, a chemically modified GTP is attached to the 5' end of the RNA.  This "cap" helps protect the RNA from degradation and helps in translation of the mRNA.
Event 2:  tailing.  At the 3' end of a gene is a sequence known as the poly-adenylation sequence (AAUAAA).  Sequence is recognized in the primary transcript and the RNA is cleaved downstream of this sequence and multiple A residues are added.  This is called the "poly-A tail".  This happens as soon as the RNA polymerase goes through the AAUAAA sequence (happens during transcription).  The polyA tail adds stability to the mRNA (immediately degraded if polyA removed).  Also required for export of mRNA from nucleus to cytoplasm.

Figure 14.12:  Third event, splicing.  The 5' end of the intron is a "splice donor" while the 3' end of the intron is a "splice acceptor".  Bind specific snRNPs which interact with one another and other proteins to form "spliceosome"--this removes introns and ligates exons back together.  Your book calls the splice donor and acceptors "consensus sequences".

Regulation of gene expression at multiple levels:  figure 14.13.

Transcriptional control for RNA polymerase II; the RNA pol that makes our mRNAs.  Figure 14.14 and 15.
Three polymerases:  I transcribes rRNA genes, II transcribes protein-encoding genes (mRNAs) and III transcribes tRNAs.  RNA is controlled by transcription factors (so are the others but different factors).  Promoter contains a sequence called TATA box.  This is just upstream of the site of initiation; RNA pol II binds here.  First TFIID, containing TATA-binding protein, binds to the TATA.  TFIIB joins and this allows RNA pol II with TFIIF to join.  More factors are needed--get transcription.

Constitutive genes--expressed all the time.  Others are controlled by regulatory sequences, either called silencers or enhancers.  Usually these sequences are found upstream of the TATA but can be in introns, downstream, or even very distant from gene.  Figure 14.15
Beta globin gene expression is regulated by 13 different proteins called transcription factors.  Not all 13 proteins are found in others cell types, so only red blood cells make hemoglobin.

Figure 14.16:  The stress response as a description of coordinated gene expression in eukaryotes.  Not found in operons like in bacteria but we do sometimes demonstrate coordinated gene expression to assure a particular response to our environment.

Transcription factors:  homemade slide for 5 families of transcription factors.  Examples of how these proteins bind DNA and interact with it are found in the book.
Figure 14.17:  the remodeling of chromatin needed for expression.

Movement of genes and gene amplification can both result in gene expression--in humans these mechanisms can be found in cancer cells.  Either the movement of the gene to a new location (chromosomal translocation) or the amplification of a gene lead to expression when it wasn't supposed to be expressed.  Results in a deregulation of the cell and cancer formation.

Figure 14.20:  Alternative splicing.

Stability of mRNA can also be controlled.  Many mechanisms exist to remove RNA from cytoplasm so that it can't continue to be translated.  This is a fast way to turn off gene expression.  Usually accompanied by a turning off of the gene but not always.

Translational and post-translational control:  translation can be controlled as well.  E.g. iron binding repressor can repress ferritin translation by binding to RNA when iron is absent.  When iron is present, we need ferritin to bind to it so repressor won't bind RNA in the presence of iron and translation occurs.

Post-translational control also at the level of protein degradation:  figure 14.21.
 
 

The essay questions on the exam will be chosen from the following questions.  Any questions with numbers from a genetic cross, specific genotypes, specific phenotypes, DNA sequences, RNA sequences or amino acid sequences will have the details changed so that memorization will not allow you to get the right answer on the exam.  You will have to know how to do the problem given to answer the question correctly.  However, the question itself will still be worded the same and will be asking for the same information as that presented here.  Studying the questions that are not on the exam as essay questions will help prepare you for the rest of the exam but you should use your book and lecture notes as well.

1.  Draw the different phases of meiosis I and meiosis II (use a diploid cell with 4 chromosomes (2 pairs) in your drawings).  Include as much written description as necessary to make the drawing absolutely clear as to what is happening.

2.  Describe 5 specific (different) ways meiosis differs from mitosis.

3.  What is the synapse and what role does it serve in meiosis and heredity?   What is a chiasmata and what is the result of chiasmata formation in heredity?

4.  What is a punnett square?  Use one to determine the genotypes from the independent assortment of the P and Y genes in the following F1 cross of pea plants:     female PpYy X male PpYy
List the different phenotypes that will result from the cross and indicate the genotypes that will result in each one.

5.  In the following imaginary crosses in plants, round seeds are dominant over wrinkled; yellow flowers are dominant over white; tall stems are dominant over short.  All three genes are linked on the same autosome.  Three crosses are made with the following result:

CROSS 1:  tall stem, yellow flowers X short stem, white flowers
    Progeny:                                                   Number
    tall stem, yellow flowers                                43
    short stem, white flowers                               45
    tall stem, white flowers                                    5
    short stem, yellow flowers                               7

CROSS 2:  yellow flowers, wrinkled seeds X white flowers, round seeds
    Progeny:                                                    Number
    yellow flowers, wrinkled seeds                        41
    white flowers, round seeds                              40
    yellow flowers, round seeds                            10
    white flowers, wrinkled seeds                            9

CROSS 3:  tall stem, wrinkled seeds X short stem, round seeds
    Progeny:                                                     Number
    tall stem, wrinkled seeds                                   47
    short stem, round seeds                                    46
    tall stem, round seeds                                         3
    short stem, wrinkled seeds                                 4

Write the genotypes next to each of the parent strains in the above crosses.
What is the recombination frequency between:
            T and Y?                                    Y and r?                                T and r?
Draw a genetic map of the part of the chromosome containing these three genes with the proper linear order and distance in map units noted.

6.  A family with inherited red/green color-blindness is analyzed.   In this family, a colorblind man married and had three children with a normal (non-colorblind) woman.  Their son was colorblind, but neither of their daughters were.  One daughter married a man from another state and had four children, two sons (one colorblind, the other normal) and two daughters, both normal.  The second daughter married and had 6 sons, none of which were colorblind.  The son married and had three daughters, none of which were colorblind.  Draw the pedigree of this family.  Indicate known genotypes (include a legend to make sure that they are understood).  Put question marks where the genotype is uncertain.  What chromosome contains the gene for red/green color-blindness in humans?

7.  In the mouse, there are three hypothetical independent loci (encoded by three different autosomes); Mc, Ke and Ms (Mc for mottled coat vs. mc for white coat, Ke for kinked ears vs. ke for smooth ears, and Ms for musty scent vs. ms for nonmusty scent).  An F1 mouse from a cross between two inbred strains, one carrying all dominant traits and one carrying all recessive traits, is back-crossed to a recessive strain mouse.  Show the genotypes of the two mice in the backcross, show the gametes each makes, and give the possible genotypes and phenotypes for the progeny, with the ratios expected.

8.  Yellow body color, encoded by the a allele, is sex-linked and recessive in Drosophila.  The wildtype allele (a+) codes for wildtype (gray) body color.  What genotypes and phenotypes (and what ratios) would you get in the following crosses?
a.  yellow bodied male with a wildtype, true-breeding female?
b.  yellow bodied female with a wildtype male?
c.  yellow bodied female with yellow bodied male?
d.  yellow bodied male with wildtype, heterozygous female?
e.  wildtype male with wildtype, heterozygous female?

9.  A plant is heterozygous at five traits, AaBbCcDdEe.  These traits assort independently.  This plant is fertilized by a plant that is AaBbCcDDEe.  Determine the expected frequency of the following progeny and show your calculations:
a.  AABBCCDDEE
b.  aabbCcddEE
c.  AaBbCCDdEe
d.  aabbCcDDEe
e.  AabbCcDdEe

10.  What are the five components of the replication complex and what is the function of each of those components?

11.  Draw a replication fork and describe what is happening at each strand.  Be sure to define the following:  the 5' and 3' ends, leading strand, lagging strand, primer(s), and Okazaki fragments.  Include as much written description as necessary to make the drawing absolutely clear as to what is happening.

12.  Draw a double-stranded DNA molecule and define the following:  the base pairing rules, the 3' end of DNA and what defines it, the 5' end of DNA and what defines it, the major groove, the minor grove, the sugar-phosphate backbone, and the hydrogen bonding, including the number, between base pairs.  You do not have to memorize the actual nucleotide structure; use a line drawing which is clear, concise and labeled very carefully to include all of the above information.  Include as much written description as necessary to make the drawing absolutely clear.

13.  Draw and/or describe the "lagging strand story" in bacteria; that is, how the Okazaki fragments are resolved and what enzymes are responsible.  Include as much written description as necessary to make the drawing absolutely clear if you use one.

14.  Define why DNA replication is semi-conservative.  Why does all DNA synthesis require a primer?  In what direction is new DNA synthesized and why?  What is DNA proofreading?  What is mismatch or excision repair and how does this differ from proofreading?

15.  Draw the process of transcription.  Include the DNA template, with the directionality noted, the production of the RNA and what is required at initiation, elongation and termination.  Include as much written description as necessary to make the drawing absolutely clear as to what is happening.

16.  What is the enzyme in the cell that is responsible for the function of the universal genetic code?  Draw and describe how this enzyme works.  Include as much written description as necessary to make the drawing absolutely clear as to what is happening.

17.  Draw the initiation of translation, indicating the order of the assembly of the parts.  Be sure to include the RNA, indicate its directionality and what signals the start of translation.  What are the four sites in the ribosome and what does each site interact with?  Include as much written description as necessary to make the drawing absolutely clear as to what is happening.

18.  Draw a ribosome on an mRNA, translating a protein.  Indicate the directionality of the template RNA and the direction of protein synthesis.  Mark the binding sites of the ribosome, indicate how each is used,  and clearly indicate how the process of protein elongation occurs.  Include as much written description as necessary to make the drawing absolutely clear as to what is happening.

19.  Draw and describe the termination of translation.  Include as much written description as necessary to make the drawing absolutely clear as to what is happening.

20.  The sequence of DNA shown below is a hypothetical gene with no introns that encodes a very small protein.  (Please note that convention states that a DNA sequence encoding a protein is always shown in the 5' to 3' direction as shown below.)  Please provide the information requested:

 5' GGCTAAATGCTTAAAAGCTACGGGCGCGAGGAATAGGAG 3'
 3' CCGATTTACGAATTTTCGATGCCCGCGCTCCTTATCCTC 5'

Which is the DNA template strand that will be transcribed?  (Write out the sequence and label the 5' and 3' ends)
What is the sequence of the mRNA that will result from this gene?  (Write out the sequence and label the 5' and 3' ends)
What is the amino acid sequence of the translated polypeptide?  (Use the universal genetic code and the three letter code.)

21.  Draw and/or describe the translation of a protein that is destined to be in the plasma membrane of a cell.  Briefly indicate how this translation differs from that of a protein that will remain in the cytoplasm of the cells.  Include as much written description as necessary to make the drawing absolutely clear if you use one.

22.  What are the four types of point mutations and their definitions?  What is the difference between a somatic and a germ-line mutation?

23.  What are the three major components of an operon and what is the function of each of these components?

24.  Draw the lac operon from E. coli and clearly indicate how lactose regulates expression from this operon.  How does glucose regulate expression from this operon?

25.  Draw the trp operon from E. coli and clearly indicate how tryptophan regulates expression from this operon.

26.  What is the function of telomeres?  Why do they shorten with age?  What is the function of telomerase and what cell types express this enzyme?

27.  In what three main ways are primary transcripts modified during "processing" and what is the purpose of the three modifications?  Define exons, introns, 5'UT, 3'UT, and splice donor and acceptor?

28.  Draw the spliceosome and describe how it splices RNA.  Be sure to label the drawing carefully so each component is clearly identified.  Include as much written description as necessary to make the drawing absolutely clear as to what is happening.

29.  Draw and describe how stress response elements, and the proteins that bind there, coregulate transcription.  Include as much written description as necessary to make the drawing absolutely clear as to what is happening.

30.  Draw and describe the function of the proteasome.  Include as much written description as necessary to make the drawing absolutely clear as to what is happening.
 
 

Part 1:  For the following multiple choice questions, enter the best single answer onto a scantron 882-ES using a #2 pencil.  (2 points each) (correct answers in bold)

1.  What is the sequence of DNA that is complementary to the following sequence?
      5' AATTGCATTCCGCGATAACGTTTCC3'
 a.  5'AATTGCATTCCGCGATAACGTTTCC3'
 b.  5'GGAAACGTTATCGCGGAATGCAATT3'
 c.  5'TTAACGTAAGGCGCTATTGCAAAGG3'
 d.  none of the above
2.  Meiosis differs from mitosis in that:
 a.  the chromosomes align on an equitorial plate during metaphase.
 b.  sister chromatids separate to different daughter cells.
 c.  the homologous pairs associate with one another in prophase I.
 d.  all of the above.
3.  Homologous recombination
 a.  occurs when sister chromatids form chiasmata, or "cross-over", in the bivalent structure.
 b.  only takes place during mitosis.
 c.  results in non-disjunction, which is the unequal distribution of DNA in daughter cells.
 d.  explains recombinant phenotypes in the inheritance of independently assorting traits.
4.  Apoptosis and necrosis differ from one another in which of the following ways:
 a.  only one of them induces inflammation and harm to surrounding "innocent bystander" cells.
 b.  only one of them results in cell death.
 c.  only one of them is triggered by conditions outside the cell.
 d.  none of the above.
5.  Two genes are linked on the same chromosome.  When two truebreeding plants, one dominant, one recessive, are crossed, 10 recombinant phenotype plants were observed in the F2 population out of a total of 600 progeny.  How many map units apart are the two genes?
 a.  0.017
 b.  0.17
 c.  1.70
 d.  17.0
6.  A hypothetical human trait, called furry toes, is seen rarely in human families.  Individuals afflicted by this trait have excess hair growth on their toes.  By studying four generations of a single family, the following pedigree was established.  Known affected individuals are indicated as having the gene and carriers are indicated when they can be inferred.

A pedigree was drawn here-learn how to read and draw pedigrees for humans.

Which inheritance pattern best explains this pedigree?
 a.  recessive X-linked gene.
 b.  dominant X-linked gene.
 c.  recessive autosomal gene.
 d.  dominant autosomal gene.
7.  Which of the following statement(s) is/are true?
 a.  Penetrance is the percent of individuals carrying an allele that display the associated trait.
 b.  Hemizygocity is extremely rare because it only occurs in individuals that have lost a chromosome.
 c.  Phenotype always reflects genotype.
 d.  none of the above.
8.  The leading strand of DNA synthesis is synthesized
 a.  as one strand from a single RNA primer located at the 5' end of the new strand.
 b.  as Okasaki fragments and requires DNA ligase to put the pieces together.
 c.  from 3' to 5' while the lagging strand is synthesized 5' to 3'.
 d.  all of the above.
9.  The lagging strand of DNA synthesis is synthesized
 a.  as one strand from a single RNA primer located at the 5' end of the new strand.
 b.  as Okasaki fragments and requires DNA ligase to put the pieces together.
 c.  from 3' to 5' while the leading strand is synthesized 5' to 3'.
 d.  all of the above.
10.  What is the significance that DNA polymerase can "proof-read"?
 a.  It reduces the error rate in DNA synthesis by 1000-fold.
 b.  It maintains the integrity of DNA during mitosis and meiosis.
 c.  It lowers the chance that mutations will be passed to new generations and to new cells.
 d.  all of the above.
11.  Which of the following statement(s) about DNA synthesis is/are false?
 a.  RNA polymerases can start synthesis de novo but DNA polymerases can not.
 b.  The start of DNA replication always occurs at origins of replication.
 c.  A primer is required only for the synthesis of the lagging strand.
 d.  none of the above.
12.  The Hershey-Chase experiment demonstrated
 a.  that DNA replicates semi-conservatively.
 b.  that DNA, not protein, is the genetic material for most organisms.
 c.  the universality of the universal genetic code.
 d.  that DNA was a double helix.
13.  RNA is similar to DNA in that they
 a.  are both always in the nucleus of a eukaryotic cell.
 b.  are both synthesized in the 5' to 3' direction by polymerases.
 c.  both are made up of deoxynucleotides.
 d.  are both single stranded.
14.  The aminoacyl-tRNA-synthetase enzymes
 a.  are responsible for the specificity of codons for amino acids.
 b.  are specific for a single amino acid.
 c.  are specific for a limited number of tRNAs determined by the anti-codon sequence.
 d.  all of the above.
15.  Transcription
 a.  is the process by which the cell makes protein.
 b.  is the process by which the cell makes RNA.
 c.   is the process by which the cell synthesizes new DNA.
 d.  all of the above.
16.  Translation
 a.  is the process by which the cell makes protein.
 b.  is the process by which the cell makes RNA.
 c.   is the process by which the cell synthesizes new DNA.
 d.  all of the above.
17.  The P site on the large ribosomal subunit
 a.  is the exit site of the ribosome where empty tRNAs leave the complex.
 b.  contains a tRNA with the growing polypeptide chain.
 c.  accepts the incoming tRNA which is bringing in the next amino acid for the growing polypeptide chain.
 d.  holds the release factor when the ribosome reaches a stop or termination codon.
18.  What are differences that can be found between proteins that will be secreted or in the plasma membrane versus proteins found in the cytoplasm?
 a.  They contain a stretch of hydrophobic amino acids at the N terminus called a signal sequence.
 b.  They are translated into the lumen of the rough endoplasmic reticulum.
 c.  They interact with the signal recognition particle which binds to a receptor on the ER.
 d.   all of the above.
19.  The open reading frame (called ORF) of an mRNA
 a.  starts with the first base in the message and goes through to the end of the mRNA.
 b.  is found in reading frame 1 of every mRNA, no matter what protein it encodes.
 c.  is the reading frame of the RNA that encodes the protein and it always begins with AUG.
 d.  does not exist because ORFs are only found in rRNAs and tRNAs.
20.  The three main processing events that precursor RNAs destined to be translated go through are
 a.  recombination, replication and splicing.
 b.  replication, capping and polyadenylation.
 c.  polyadenylation, splicing and recombination.
 d.  capping, polyadenylation and splicing.
21.  The spliceosome is responsible for
 a.  transport of the rRNAs from the nucleus to the cytoplasm where they can be assembled into ribosomes.
 b.  the removal of introns from the precursor RNA to make the mature mRNA.
 c.  the removal of exons from the precursor RNA to make the mature mRNA.
 d.  the attachment of the amino acid to the tRNA for protein synthesis.
22.  RNA polymerase II is responsible for the transcription of which type of RNA?
 a.  mRNAs
 b.  tRNAs
 c.  rRNAs
 d.  all of the above.
23.  Which of the following statements about polyadenylation is/are true?
 a.  Polyadenylation is the addition of ~200-300 A residues at the end of an mRNA.
 b.  Polyadenylation occurs at a specific sequence, AAUAAA, in the 3'UT of an pre-mRNA.
 c.  Polyadenylation is required for mRNA stability.
 d.  all of the above.
24.  Pseudogenes
 a.  are just like regular genes because they encode proteins.
 b.  always have intron like sequences just like real genes.
 c.  often have intron sequences removed, leading to the theory that they arose from mRNAs.
 d.  are not found in the human genome.
25.  Which of the following statements concerning telomers is/are true?
 a.  Telomers are made up of a repetitive DNA sequence, 5'TTAGGG3', that is repeated hundreds of times at the end of each chromosome.
 b.  Telomers protect us from losing vital chromosomal DNA during DNA replication.
 c.  Telomers are lost in cells that don't express telomerase, limiting the number of cell divisions a cell can undergo.
 d.  all of the above.

Part 2:  For the following true-false questions, enter (A) for true and (B) for false onto a scantron 882-ES using a #2 pencil.  (1 point each)

26.  The synapse is a physical association of the two chromosomes in a homologous pair during meiosis.  T
27.  Mitosis maintains genetic material while meiosis promotes genetic diversity.  T
28.  Apoptosis is programmed cell death in which the cell degrades its DNA and signals phagocytes to remove it by phagocytosis.  T
29.  The definition of epistasis is that the trait caused by a single gene can be influenced by the alleles present at other loci.  T
30.  The ABO blood group system contains multiple alleles, two co-dominant and one recessive.  T
31.  DNA polymerase reads the template strand 3' to 5' to synthesize the new DNA strand 5' to 3'.  T
32.  Mismatch repair is a mechanism in which several base pairs of DNA are removed from one strand and replaced with the correct nucleotides reading the other strand.  F
33.  Point mutations always result in the substitution of amino acids in the final protein product of a gene.  F
34.  Alu sequences, a moderately repetitive DNA sequence, may function as the origin of replication in human DNA.  T
35.  A somatic mutation is passed on to the next generation due to the fact that it took place in cells giving rise to eggs and/or sperm.  F

Part 3:  Answer the following questions as indicated.  10 points each.

36.  Draw the lac operon from E. coli and clearly indicate how lactose regulates expression from this operon.  How does glucose regulate expression from this operon?

37.  The given sequence of DNA is a hypothetical gene with no introns that encodes a very small protein.  Provide the information requested:

 5' GGCTAAATGCTTAAAAGCTACGGGCGCGAGGAATAGGAG 3'
 3' CCGATTTACGAATTTTCGATGCCCGCGCTCCTTATCCTC 5'

Which is the template strand?  (Write out the sequence and label the 5' and 3' ends)
What is the sequence of the mRNA?  (Write out the sequence and label the 5' and 3' ends)
What is the amino acid sequence of the translated polypeptide?  (Use the attached universal genetic code and use three letter code.)

38.  In the mouse, there are three hypothetical independent loci (encoded by three different autosomes); Mc, Ke and Ms (Mc for mottled coat vs. mc for white coat, Ke for kinked ears vs. ke for smooth ears, and Ms for musty scent vs. ms for nonmusty scent).  An F1 mouse from a cross between two inbred strains, one carrying all dominant traits and one carrying all recessive traits, is back-crossed to a recessive strain mouse.  Show the genotypes of the two mice in the backcross, show the gametes each makes, and give the possible genotypes and phenotypes for the progeny, with the ratios expected.

39.  What are the five components of the replication complex and what is the function of each of those components?