BIOL 1090 DNA & Chromosomes

Functions of genetic material genotypic function is replication: Parental DNA separates and new complementary strands synthesize creating two identical daughter DNA molecules phenotypic function is gene expression. Phenotype is defined as how function is read from expressing gene. Phenotypic function: how genome is used; expression of individual gene; genes silent/not at right times; want to pass on genes (success or failure according to nature if you can pass on genes or not) evolutionary function is mutation: population variation is the substrate on which natural selection acts. There is variation in every population (spots on ladybugs). There can be an innocuous allele but if the environment changes and the allele is favoured then it may become expressed, and it can even make a whole new species if enough mutation and expression occurs. Myostatin: regulation of muscle mass size. MET 5 not in all cells, only expressed in ones that will be muscle. Cells keep dividing until they become more specialized Genomes are not deterministic: genomes can interact with the environment such as the Daphnia which can either be long or short depending on the situation. Daphnia on left is longer and thus harder for predators to eat while the one on the right had the absence of predator signal so developed looking like a water flea. At the heart of genes and genetics is DNA (Deoxyribonucleic acid), which is a polymer. Consists of a phosphate group, a 5 carbon sugar (2­deoxyribose), and one of 4 cyclic nitrogenous bases. Single are C and T and are pyrimidine while double are A and G which are purine. Deoxythymidine monophosphate (dTMP) Deoxycytidine monophosphate (dCMP) Deoxyadenosine monophosphate (dAMP) Deoxyguanosine monophosphate (dGMP) Purine and Pyrimidine nucleotides in polynucleotide chains are connected by phosphodiester bonds. DNA is chemically polar, 5’ end has a free phosphate group while 3’ end has a free hydroxyl group

DNA is double stranded and the strands are antiparallel so 5’ end next to 3’ end of opposite strand. The double helix is right handed, meaning strands twist clockwise when holding it away from your body and looking at it. Strands held together by hydrogen bonds between bases on opposing strands and hydrophobic interactions between stacked bases. Longer DNA is more stable since more bonds and more stabilizing hydrophobic interactions. Base pairing is specific and mediated by H bonds. Opposing strands are complementary (A always with T ; C always with G) CG more stable than AT since CG has 3 H bonds while AT only has 2 H bonds. Most common form of DNA is B­DNA, while A­DNA also exists. Z­DNA is the one with left handed twist. There is a minor and major groove in DNA (there is alternating narrower and wider parts in DNA). This is for protein binding to DNA since it can recognize the different grooves and bind where needed. Eukaryotic chromosomes are composed of proteins, DNA, and RNA. Almost 2/3 of nucleus is DNA in chromatin form. There is lots of protein which are called histones, but also nonhistone proteins see slide First level of condensation is packaging DNA as a negative supercoil into nucleosomes, producing an 11 nm fibre. Nucleosome core is 146 nucleotide pairs of DNA wrapped 1 and ¾ turns around an octamer of histones. The linker DNA joining these cores is varying in length from 8 to 114 nucleotide pairs. Core nucleosome made of 8 histones, 4 types and 2 of each, 2 H2a, 2 H2b, 2 H3, 2 H4. 9th histone anchors the nucleosome, histone H1 The second level of condensation ­ an additional folding or supercoiling of the 11 nm fibre to produce a 30 nm fibre. Driven by nucleosomal interactions. Histone H1 is involved. The two models describing the substructure are the solenoid and the zig­zag models. See slide 24 Lecture 1. Models can be drawn differently depending on how they are viewed but it is still the same DNA in the 30 nm fibre. The solenoid model is favoured as to how DNA looks in a living cell. The third level of condensation is the attachment of the 30 nm fibre at many positions to a (non­histone) protein scaffold. In metaphase the chromosome uses the 30 nm fibre as the basic structural unit because DNA is in its most condensed form. The protein scaffold lets the fibre loop to reduce even more space. Scaffold then turns in helix shape to continue reducing space.

Prokaryotic chromosomes are also highly compact, being folded and supercoiled reducing the original circular shape and size from 350 um to 2 um. DNA found in mitochondria and chloroplasts exists in circular chromosomes that resemble those of prokaryotes (still has coiling). (Right handed helix means left handed twist to form supercoils). Endosymbiotic: 2 different organisms where one cell lived inside the other and this eventually evolved in eukaryotic cells ­ now the inside organism NEEDS the outside one to function and survive (evolved into dedicated organelle). Chromosomes ends are protected by telomeres. Specialized sequence of base pairs 5’ 1­4 T , 0­1 A , 1­8 G 3’. One example would be TTAGGG. Telomeres provide three important functions: 1. Resist degradation by DNAses ­ cell has enzymes that can degrade DNA such as endonuclease or exonuclease. 2. Prevent fusion of chromosomal ends ­ free ends of chromosomes can fuse together and this would want those molecules to stick together since phosphodiester bonds will be formed so telomeres prevent this fusion. 3. Facilitate replication of the ends of the linear DNA ­ Every time cell divides the chromosome gets a bit shorter and since telomeres have thousands of repeats they provide a buffer for the loss of the DNA at the end. Telomeres get shorter at the end of every cell division but they are long enough that this does not matter. Telomerase keeps telomeres long , but cancer cells also have this to keep dividing and spreading Another feature of chromosomes is centromeres: the waist where the chromosome constricts. They provide the point of attachment of chromosomes to microtubules in the mitotic spindle. Centromeres also have specific base pair sequence. AT sequences have 2H bonds so more flexible than average sequence → can bend DNA more tightly. Yeast centromeres are 110­120 base pairs (bp) long. Three essential regions: regions I and III are conserved sequences that bind proteins involved in spindle attachment. region II ~90 bp, >90% A, T Centromeres in multicellular eukaryotes are much larger and more complex, eg 5000­15000 copies of the 171 bp alpha satellite sequence. Binding sites for a protein called CENP (related to Histone H3). Cell Cycle & Mitosis

Mitosis Cellular organelles and cytoplasmic contents are divided more or less equally between daughter cells. The endoplasmic reticulum and Golgi complex are fragmented at the time of division and reformed in the daughter cells. Mitochondria and chloroplasts are randomly divided between daughter cells; however chromosomes must be duplicated exactly and distributed equally and exactly to daughter cells Stages of Cell division Cell cycle is composed of Interphase and mitosis Interphase is broken into G1, S, and G2 phase Mitosis is broken up into Prophase, Metaphase, Anaphase, Telophase and ends with cytokinesis. Mitosis is also called the M phase in the cell cycle G1 phase (Gap 1); growth, cellular metabolism S phase (Synthesis); DNA replication (chromosome duplication) G2 phase (Gap 2) preparation for mitosis M phase(mitosis); chromosomal ‘pas de deux” and cytokinesis Interphase ­ the time between successive mitoses (G1+S+G2) ** not all cells undergo mitosis some leave during G1 and enter G0, they are said to be quiescent. ** no “clock” that regulates cell cycle timing in eukaryotic cells Centrosome cycle is when the centrioles are duplicated, occurs along the cell cycle In animal cells, the centrosomes are microtubule organizing centres (MTOCs) negative end of MTOCs go towards centrosomes and positive ends go away from centrosomes, this orientation is key Chromosomes duplicate at the start of mitosis Duplicated chromosomes at metaphase condense under the influence of condensin Cohesin holds strands of DNA together during interphase, during prophase condensin pinches strands of DNA forming loops in the strands, these loops are then adhered together by cohesin. See slide 10 for picture n = haploid chromosome number c = DNA content

Diploid mother cell(2n)(4c) → DNA synthesis and then divides (Fission) → Two daughter cells (2c)(2n) ** they are identical diploid cells Interphase­ when mitosis begins each chromosome has been duplicated, producing sister chromatids joined at the centromere by cohesin, the centrosome is duplicated Prophase ­ initiation of spindle formation, condensation of duplicated chromosomes, fragmentation of ER and Golgi, nucleolus disappears, Nuclear membrane breaks down, spindle microtubules invade the nuclear space. Prometaphase ­ chromosomal microtubules attach to the kinetochores, located on the outer surface of centromeres, chromosomes move towards the equator of the spindle. Kinetochores ­ outer kinetochore microtubule binding and microtubule motor activity signal transduction, inner kinetochore centromere replication chromatin interface kinetochore formation. Positive end of microtubules where subunits are added or lost. Other end is negative. Metaphase ­ duplicated chromosomes are aligned midway between the spindle poles, this equatorial plane is called the metaphase plate. Anaphase ­ centromeres split and chromatids separate chromosomes move towards opposite spindle poles, spindle poles move further apart. Telophase ­ chromosomes cluster at opposite spindle poles, chromosomes become dispersed and decondense, nuclear envelope assembles around chromosomes, Golgi and ER reform, daughter cells form by cytokinesis In plant cells, a cell plate forms the scaffold for a new, cellulose­containing, cell wall Meiosis & gametogenesis n = number of chromosomes = the haploid state 2n chromosomes = the diploid state Homologues ­ similar chromosomes that are similar to one another but not identical (sister chromosomes) Heterologues ­ chromosomes that are not identical or similar (chromosomes that do not belong to the same pair) Sex chromosomes ­ Specialized chromosomes, in female humans it is XX and in male humans it is XY. Meiosis I (Leptonema, Zygonema, Pachynema, Diplonema); chromosomes, each consisting of two sister chromatids, begin to condense. Homologous chromosomes begin to pair,

homologous chromosomes are fully paired, homologous chromosomes separate, except at chiasmata Pairing of homologous chromosomes is called synapsis and is often facilitated by formation of a synaptonemal complex; composed of lateral elements connected to a central element by transverse fibers. When homologous chromosomes are fully paired, paired sister chromatids are called Tetrads while paired sister chromosomes are called Bivalents. Crossing over involves the breakage of chromatids and the exchange of the broken pieces between homologous chromosomes (non­sister chromatids). Following crossing over, homologous chromosomes start to pull apart, but remain joined at the cross over junctions (called chiasmata). **Diplotene stage may persist for the entire reproductive life of the individual (more than 40 years in female humans) Cell division in Meiosis I is called reduction division Prophase I: Paired chromosomes condense further and become attached to spindle fibers. Metaphase I: Paired chromosomes align on the equatorial plane in the cell. Anaphase I: Homologous chromosomes disjoin and move to opposite poles of the cell. Telophase I: Chromosome movement is completed and new nuclei begin to form. ** Chromosome disjunction occurs between metaphase I and anaphase I Meiosis I produces two haploid daughter cells that are genetically identical Meiosis: Diploid(2c, 2n, meiocyte)­­DNA synthesis→ 4c ­­Division 1→ Two 2c cells ­­ Division II → four c cells (they are nonidentical haploid (n) cells). Meiosis II Prophase II: Chromosomes, each consisting of two sister chromatids, condense and become attached to spindle fibers. Metaphase II: Chromosomes align on the equatorial plane in each cell Anaphase II: Sister chromatids disjoin and move to opposite poles in each cell (chromatid disjunction occurs) Telophase II: Chromosomes decondense and new nuclei begin to form. Cytokinesis: The haploid daughter cells are separated by cytoplasmic membranes Kinetochore position changes between prophase I and II

Spermatogenesis Spermatogonia (2n); Mitosis occurs for several generations → chromosome duplication leads to Primary spermatocyte (2n) ­­Meiosis I→ Secondary spermatocyte (n) ­­Meiosis II→ spermatids attached to spermatocytes which are interconnected by cytoplasmic bridges → four sperm cells(n) Oogenesis Oogonium (2n) ­­Chromosome duplication and pairing→ primary oocyte ­­Meiosis I→ Primary oocyte dividing → First Polar body and Secondary oocyte First polar body ­­meiosis II→ polar body dividing → Two polar bodies(n) Secondary oocyte ­­secondary oocyte dividing→ polar body(n) and egg(n) **Meiosis I begins in the fetal ovary but arrests at Prophase I (at diplonema) ** For ovulation, an egg “matures” to metaphase II and is released into the fallopian tube. At this time of fertilization the oocyte is still 2c and meiosis II is completed(expulsion of the second polar body) after fertilization Baker’s yeast; fate of a haploid cell Baker’s yeast → Asexual “budding” → Haploid cells of opposite mating types ­­mating→ Diploid zygote → Meiosis → sporulation → Ascus with four haploid ascospores → Baker’s yeast Sporophyte (diploid) GAmetophyte (haploid) A true “alternation of generations” in plants Sporophyte → male and female germ cells → MALE meiosis leads to microspores while FEMALE meiosis leads to megaspores → Mega and microspores go through mitosis and become MALE and FEMALE gametophytes → gametes ­­fertilization→ zygote (zygote microspore) ­­Mitosis→ Sporophyte The basic principles of Mendelian inheritance The benefits of Mendel’s experimental system: his peas were highly in­bred. Because flower structure promoted self­fertilization they therefore bred true. Eg when two tall plants were crossed, they only produced tall progeny. Mendel’s experiments were designed so that he could study one trait at a time.

Monohybrid cross (single trait is followed). In a genetic cross, the parents are referred to as the parental (P) generation. Their offspring represent the first filial (F1) generation. Their grand­offspring are the F2 generation, etc. 1. Tall and dwarf varieties are cross­fertilized. 2. All the hybrid progeny are tall. 3. The hybrid progeny are self­fertilized. 4. Tall and dwarf plants appear among the offspring of the hybrids approx in a ratio of 3 tall : 1 dwarf. This ratio found for seven other traits tested. Each trait appeared to be controlled by a heritable factor that came in one of two forms: dominant and recessive. Mendel’s heritable factor = gene dominant and recessive forms = alleles Homozygous ­ both alleles are identical Heterozygous ­ the two alleles are different Centromere not exactly in the middle of a chromosome, there is a short or p arm and a long or q arm. Locus (plural loci) is a fixed position on a chromosome. A gene or allele will be found on this position. Tall is dominant and dwarf is recessive allele in plants.

(1) Each parental homozygote produces one kind of gamete. P generation Tall DD x Dwarf dd produce one D gamete and one d gamete.

(2) The F1 heterozygotes produce two kinds of gametes in equal proportions. F1 generation Tall Dd produces one D and one d gamete.

(3) Self­fertilization of the F1 heterozygotes yields tall and dwarf offspring in a 3:1 ratio. F2 GENERATION (SELF FERTILIZATION)

D d

D Tall DD Tall Dd

d Tall Dd Dwarf dd

Phenotypes Genotypes Genotypic ratio Phenotypic ratio

Tall DD 1 add 1 to 2 to get 3

Tall Dd 2 3

Dwarf dd 1 1

Mendel’s Principle of Dominance: in a heterozygote, one allele may conceal the presence of another. Mendel’s Principle of Segregation: neither allele is typically changed by coexisting with other in a heterozygote. Two different alleles segregate from each other during the formation of gametes. In Anaphase I the homologous chromosomes disjoin and move to opposite poles of the cell. Maternal allele one pole and paternal allele to the other pole. Dihybrid cross ­ are two traits inherited independently? P Yellow, round x Green, wrinkled F1 Yellow, round → self fertilizes F2 Yellow, round / Green, round / Yellow, wrinkled / Green, wrinkled

315 108 101 32 Approx 9:3:3:1 ratio Yellow, round appear to be dominant traits. 4 Phenotypic classes in the F2 generation represent all possible combinations of the colour and texture traits: 2 that resemble the parents plus 2 new trait combinations. The ratio of phenotype classes suggested to Mendel a scenario in which each trait was controlled by a different gene, each segregating two alleles. 1. Each parental homozygote produces one kind of gamete. P Yellow, round (GG WW) x Green, wrinkled (gg ww) Gametes GW/GW/GW/GW gw/gw/gw/gw 2. The F1 heterozygotes produce four kinds of gametes in equal proportions. GW x gw F1 Yellow, round with Gg Ww and gametes are GW / Gw / gW / gw Mendel’s Principle of Independent Assortment: the alleles of different genes assort/segregate independently of each other. In Anaphase I ­ g/G and G/g move to opposite poles while W/w and w/W also move to opposite poles (homologous chromosomes).

3. Self fertilization of the F1 heterozygotes yields four phenotypes in a 9:3:3:1 ratio

GW Gw gW gw

GW GGWW GGWw GgWW GgWw

Gw GGWw GGww GgWw Ggww

gW GgWW GgWw ggWW ggWw

gw GgWw Ggww ggWw ggww

9 Yellow, round : 3 Yellow, wrinkled : 3 Green, round : 1 Green, wrinkled (Ww is the same as WW since it results in both being Round). Pedigrees and human genetics For non­humans, we can deduce the genotype by breeding Since a heterozygote may have the same phenotype as the homozygous dominant, a testcross may be performed to determine the individual's genotype. In a test cross, the individual of unknown genotype must be crossed with a homozygous recessive individual. eg. a testcross involving a plant of the genotype DD Gg Ww would involve crossing this plant to one with the genotype dd gg ww. Pedigrees show the relationships between members of a family. A trait is likely showing a dominant mode of inheritance if: every affected individual has at least one affected parent the trait is manifested in at least one individual in every generation once the trait appears (the spontaneous appearance of a dominant allele is extremely rare)

A trait is likely showing a recessive mode of inheritance if: the trait suddenly appears in a pedigree the trait skips a generation In the absence of evidence to the contrary, assume that unrelated individuals marrying into the family do not carry the recessive allele Human families are relatively small, therefore phenotypic ratios among offspring often deviate significantly from Mendelian expectations. Consider a couple, each heterozygous for a recessive allele that causes a serious disease in homozygous individuals. If they have four children, what is the probability that exactly one is affected? 5 possible outcomes and multiple ways of arriving at some of them

1. 4U, 0A UUUU 2. 3U, 1A AUUU, UAUU, UUAU, UUUA 3. 2U, 2A UUAA, UAAU, AAUU, UAUA, AUAU, AUUA 4. 1U, 3A UAAA, AUAA, AAUA, AAAU 5. 0U, 4A AAAA

The probability of all possible outcomes can be calculated: **Treat each birth as independent** P(affected) = P(cc) = ½ x ½ = ¼ P(Unaffected) = P(CC) + P(Cc) = ¼ + ½ = ¾ Unaffected Affected 4 0 1 x (¾) x (¾) x (¾) x (¾) = 81/256 3 1 4 x (¾) x (¾) x (¾) x (¼) = 108/256 2 2 6 x (¾) x (¼) x (¼) x (¼) = 54/256 1 3 4 x (¾) x (¼) x (¼) x (¼) = 12/256 0 4 1 x (¼) x (¼) x (¼) x (¼) = 1 /256 For a total number of n progeny we can calculate the Binomial probability that exactly x progeny will fall into one class, and y as the other class as :

[n!] / [x!y!] pxqy where the two classes are P and Q, with probabilities of occurrence of p and q. Since there are only two classes p =1 ­ q Solution to previous problem: [4!]/[3!1!] (¾)3(¼)1 = 4 x (27/64) x (¼) = 108/256

1. From the pedigree, albinism appears to be a recessive trait 2. S is therefore homozygous recessive for a 3. R has siblings with albinism, so there is some risk that R is a carrier (heterozygous Aa) 4. The risk of T having albinism therefore depends on two factors: the probability that R is a

carrier and the probability that R transmits the A allele to T if he is. Among the offspring without albinism ⅔ are heterozygotes Risk calculation that child T has albinism: PT(aa) = PR(Aa) x PR(a) x PS(a) = ⅔ x ½ x 1 = 2/6 = ⅓ The absence of a phenotype doesn’t necessarily reflect the absence of a causative genotype. A dominant trait like nonpolypoid colorectal cancer that may not manifest until after reproductive age makes risk assessment difficult

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I must be a heterozygote because her mother is homozygous and therefore the chance of her passing on the recessive allele is ½ The parents of H must both be heterozygous since they have an affected daughter, but aren’t themselves affected H therefore has a 2/3 probability of being a carrier and, if he is a carrier, the chance of him passing on the recessive allele is ½ The probability that the child will have the trait is ½ x 2/3 x ½ =2/12 =1/6 Extra Class notes: If the man was homozygous dominant then all of his offspring would have the trait since it would be passed on. Affected individuals can’t be homozygous dominant because the offspring are not carrying the trait and have to look at mother/father. U is unaffected and A is affected 1 x (¾) x (¾) x (¾) x (¾) since there is only on combination next line multiplied by 4 since there is four possible outcomes than can happen third line is times 6 since there is six possible outcomes etc Get 42% that 1 of 4 children will have cystic fibrosis if both parents heterozygous. in the equation n is number of children (4), x and y are two outcomes (x is unaffected and y is affected in this case) p is chance of being affected and q is chance of being unaffected

Recessive trait since neither parent has it (for sliding asking for risk that R and S will have a child with albinism) R might be Aa but may also be AA Chances that R is heterozygous is ⅔ BECAUSE it does not have albinism so it is not ½ it is in fact ⅔ Allelic Variation and Gene Function Lecture notes: Outside of Mendel’s model, things are not so simple; genes may (and usually) have more than 2 alleles, different alleles may affect the phenotype in different ways, A single gene may control several traits, Multiple genes may control a single trait. Most common allele is considered the wild type ( designated with a superscripted +, A+ or c+) any allele found at appreciable frequencies, at least 1%, in the population is called a polymorphism. All other alleles are considered mutant. Recessive mutations almost involve a loss of function; null allele (complete loss of function), hypomorphic allele (partial loss of function). Loss of function is usually recessive because most gens only require one functional copy for normal(wild type) phenotype. Different Types of Mutations Dominant mutations: can involve a loss of function OR a gain of gene function. Loss of function: for genes in which one functional copy is not enough, this is called haploinsufficiency Dominant negative; a loss of function mutation that interferes with the normal function of the wt allele Gain of function; enhances the normal function of the gene Antennapedia mutation in Drosophila is a (dominant) gain of function mutation in the regulatory region of the Antp gene Incomplete Dominance Phenotype of the heterozygote is in between the two phenotypes of the two homozygotes, one allele is either partially or incompletely dominant over the other. E.g. Red and white flower producing a pink flower. precursor + polypeptide W (Gene W) → product (red pigment) → phenotype

Red → WW → 2x Pink → Ww → x White → ww → 0 Codominance The heterozygote expresses phenotype of both homozygotes, neither allele is dominant Dominance Hierarchy Allelic series describes the dominance hierarchy of multiple alleles Example: Wild type (c+c, c+cch, c+ch) > Light chinchilla (cch c) >= Light chinchilla with black tips (cchch) > Himalayan (ch c). c+ > cch >= ch > c For allelic series for human blood IA and IB are codominant over i Single Gene affecting several traits AY: a dominant visible allele that is also a recessive lethal. Example: Heterozygous AYA+ cross produces 1 A+A+ gray brown (agouti) rat, 2 AYA+ yellow rat, and 1 AYAY embryonic lethal. Multiple genes affecting a single trait Gene A → Polypeptide A combines with Polypeptide B (from Gene B) → wild type phenotype Testing whether two alleles that confer a particular phenotype are in the same gene is called a complementation test. Example: Complementation test for allelism involving recessive eye colour mutations in Drosophila. Cross of cinnabar & scarlet when they are different recessive alleles in the same gene since they only produce the same offspring. Cross of cinnabar & scarlet when they are recessive alleles in different genes produce several different combinations. By crossing a new recessive mutation with several tester genotypes you can produce several hybrid phenotypes, from the ones that are mutant you can conclude that the tester genotype and new recessive mutation are alleles. substrate + Cinnabar protein (from cinnabar gene) → intermediate + scarlet protein (scarlet gene) → brown pigment → red/brown (wf) eyes

Key points Gene action is affected by biological and physical factors in the environment. Two or more genes may influence a trait. A gene is pleiotropic if it influences multiple aspects of an individual’s phenotype. Class notes: Most genes known are multifunctional Wild type has nothing to do with dominant/recessive, just shows frequency (it is most common) 1% is an arbitrary number for the frequency for which a allele is considered polymorphism Polymorphisms were at one time mutants Example of null allele is something that prevents gene from being transcribed/protein transcribed but not translated = complete loss of function Example of hypomorphic allele is something like an enzyme, it slows down the process of an enzyme. Not optimal. Mitf +/+ has 8 while Mitf +/­ has 4 since it has only half the dominant allele but the amount required is still enough to function sufficiently. One of the genes that could lead to albinism. ­/­ would have no function for Mitf. Mitf +/M has protein­protein binding domain but doesn’t posses the replication domain (many of the replicated proteins are incomplete), Has two completed proteins. This is a dominant negative mutation.

Fly has legs growing out of head and it is gain of function. Regulatory mutation (most common) so the thorax that produces antenna instead produces legs. Gene W makes protein and then enzyme which makes pigment that is deposited into petals of flowers. WW makes 2x the amount so fully red, Ww makes 1x the amount so pink, and ww makes none so white. Incomplete dominance. Blood type M contains mostly anti­M serum with very little anti­N serum. Blood type M N contains both antiserum M and N. Blood Type N contains mostly antiserum N. Codominance. Wild type allele is the most dominant in the dominance hierarchy, and will always appear wild. Gene A can by synthesizer and then B is transport or both can be used to make enzymes and then make stuff. So single phenotype with multiple genes has many different things that the genes can do. Crossing cinnabar and scarlet produces cscc which leaves no wild type and only mutant. So cinnabar and scarlet are on same gene (no complementation). Cinnabar and scarlet on different genes so cc SS and CC ss. All possible outcomes leave heterozygous so the mutations compliment each other and thus all the offspring are wild type. ** All or nothing effect; either all the offspring are wild type or all the offspring are mutant. Mutations and their effects(shit happens) Side notes: Mutations have the ability to arise spontaneously as a result of an error within DNA synthesis. Such errors include the inherent fallibility of replication proteins, as well as the incorporation of rare isoforms of the four nitrogenous bases that have had their pairing characteristics altered. Nitrogenous bases of DNA exist in two isoforms (tautomers) Common form of Thymine (Keto form); 4th carbon has double bond to oxygen and nitrogen in the third position with three hydrogens bonded to it. Rare form of thymine ( Enol form); Carbon in the fourth position is now bonded to an alcohol group and now there is a double bond between the nitrogen in the third position and the carbon in the fourth position.

Common form of Cytosine (Amino form); Carbon in the fourth position is bonded to amino functional group and double bonded to the nitrogen in the third position. Rare form of Cytosine ( Imino form); Carbon in the fourth position now double bonded to imino group and the nitrogen in the third position is bonded to three hydrogens ( ammonium?). Common form of Adenine ( Amino form); Carbon in position six has an amino functional group and is double bonded to the nitrogen in position one. Rare form of Adenine (Imino form); Carbon in position six has imino functional group and is no longer double bonded to the nitrogen in position one. Common form of Guanine (Keto form); Carbon in position six has a ketone functional group and is single bonded to the nitrogen in position one. Rare form of Guanine ( Enol form); Carbon in position six has an alcohol functional group and is double bonded to the nitrogen in pos ** Tautomers allow for non regulatory altered base pairing, i.e. Cytosine to Adenine and Thymine to Guanine. Incorporation of rare isoform during DNA replication leads to mutation in DNA sequence. e.g. Rare enol tautomeric form of guanine(G*) can lead to a mutant progeny being produced during the second­generation. Somatic mutations of this kind may affect the individual in which they occur but will not be passed on to future progeny. Increase in mutations within the p53 codon leads to increased risks of developing cancer, however the frequency at which these mutations occur cannot be predicted. Germline mutations will be inherited (mutations that occur during meiosis such as during oogenesis and spermatogenesis) “Hot spots” for spontaneous mutations during DNA replication simple repeats; repeat of a nucleotide ­ A A A A A A A A A A Direct repeat of a dinucleotide ­ G C G C G C G C Direct repeat of a trinucleotide ­ T A C T A C T A C T A C Symmetrical repeats; complementary base pairing within DNA strand ( this creates a hairpin loop within the DNA) Palindromes; G A A T T C C T T A A G Mutations can be induced by exposure to chemical mutagens

chemical mutagens can be divided into two groups: only mutagenic to replicating DNA (e.g. base analogues, acridine dyes) Mutagenic to both replicating and nonreplicating DNA (e.g. alkylating agents) ­ ethyl methanesulphonate (EMS) Mutations can also be induced by exposure to radiation, any wave length greater than the visible spectrum gives off radiation. Cosmic rays give off the most radiation. Adsorption of UV energy by pyrimidines results in their dimerization. Two thymine bases exposed to UV light combine together and form a Thymine dimer ( the double bonded carbons found within each molecule bond to the opposing thymines double bonded carbons). The thymine dimer causes a ‘kink’ with the DNA strand Each amino acid is specified by one or more codon There are no spaces between codons, codons are adjacent Genetic code is non­overlapping, each nucleotide is part of one codon The genetic code is degenerate, most amino acids are specified by more than one codon The genetic code is (nearly) universal, with minor exceptions, each triplet/codon has the same meaning in all organisms. Single base mutations; AGA(Arginine) → AGG (Arginine); silent mutation AGA(Arginine) → TGA (‘stop’); nonsense mutation AGA(Arginine) → AAA (lysine); conservative missense mutation AGA(Arginine) → AGT (serine); nonconserved missense mutation Single base changes in the beta globin gene causes sickle cell anemia; this mutation causes abnormal aggregation of hemoglobin molecules. Acridines intercalate between adjacent base pairs and distort the double helix, when these molecules replicate, additions and deletions of one to a few base pairs occur. Base pair insertion and deletion can completely change the codon sequence. Mutations can also be induced by DNA itself ­ transposable elements this element jumps into the middle of a gene causing an elongated mRNA protein ergo producing an inactive truncated polypeptide. At least 15 human inherited disorders result from expanding triplet/trinucleotide repeats

usually < 40 copies of triplet repeat are stably inherited, larger numbers of copies are unstable and prone to expansion. E.g. Fragile X syndrome (CGG), Huntington disease (CAG) Expanding triplet/trinucleotide repeat diseases can show increased severity and/or earlier onset from one generation to the next. See slide 22 for picture Lecture notes: DNA polymerase is an imperfect protein, it makes mistakes when copying DNA. Tautomers in constant motion and go back and forth between configurations but spend much more time in the normal form. Altered base pairing such as AC and GT. p53 acts in between G1 and S phase, almost all cancers have mutation with this p53 gene. Even though the mutation rate is constant across this gene, there are certain position that when mutated are linked to increased risk of cancer. For symmetrical repeats the symmetrical repeats have the complementary base pairs on the SAME strand of DNA so they sometimes fuse together. It is inverted so they can fuse together. Thymine dimers are two T’s on the same strand and the UV can cause covalent bonds on the two T’s making them dimerized and thus produces a kink in the double helix. AAA similar chemically to AGA so have similar effects on protein so it is conservative mutation, only a slight difference that won’t have a massive impact on the protein. AGT not similar to AGA, they are different chemically so it is non­conservative and thus there will be a difference to the protein. Nonsense mutations effect depends on where along the string of codons it appears. beta globin comes together in terric group to form hemoglobin. Missense mutation in DNA for beta globin can cause improper folding thus not allowing the hemoglobin to form properly, this leads to sickle cell anemia since the hemoglobin can not properly carry oxygen. Fragile X is the most common form of retardation in people

There are many ways to lose protein function ­ II, mutations that affect non­coding regions; 1) to prevent or reduce transcription. 2) Prevent translation ­ mRNA is unstable, ribosomes can’t bind, mutation of the start codon, nuclear export of mRNA is compromised. Myotonic dystrophy type I ­ CTG expansion in the untranslated region (UTR) of an mRNA transcribed from chromosome 19 encoding an enzyme (kinase). Myotonic dystrophy type II ­ CCTG expansion in an intron of a gene on chromosome 3 encoding a zinc finger transcription factor. *In both cases, the mRNA may become too large for efficient export to the cytoplasm. Sex Chromosomes and Heredity In class notes: Sex ( ° ʖ °) chromosomes vary between different species Bridge says X chromosomes fail to disjoin in anaphase I during meiosis. So females can make haploid cells that retain BOTH X chromosomes. Left side of chart on slide 16 is normal outcome and then right side is the exceptional outcome. He got XXX and XXY which led to white eyed females and XO red eyed male and YO male which dies. *O stands for nullo­x which means there was no chromosome given to it and it dies. Therefore gene for eye colour is on the X chromosome. 1/16 chance for ? to have hemophilia. (slide 22) Flies boost so they have 2 X chromosomes. Mammals inactive 1 X. C. elegans reduce function of both by half to get one X these mechanisms equalize the number of x chromosomes X inactivation is the mechanism by which we equalize gene dosage on the chromosome. Which X is deactivated is random between individuals and it is random cell by cell. Heterozygous pigment cats have orange or brown patches of fur due to the difference in which X allele is equalized since one has brown fur and one has orange (randomness in X inactivation). Barr bodies usually on edge of nucleus, associated with nuclear envelope. Can have one barr body in XX female or even 2 barr bodies in an XXX female.

Gene Structure and Transcription: Lecture slide notes: Lecture Slide notes: The transfer of information from DNA to protein is a two step process in all organisms. Central Dogma; DNA (gene) ­­Transcription→ mRNA ­­Translation→ Protein Definition 1: A gene is transcribed region of DNA Gene encode one of several types of RNA Transcription: Gene regions of DNA ­­Transcription→ RNA transcript processing → snRNA (goes to spliceosome which affects RNA) or rRNA or tRNA or mRNA or Pre­miRNA → exits through nuclear pore. Translation: Pre­miRNA → dicer → miRNA → cleavage of mRNA OR repression of translation mRNA → Ribosome → production of polypeptide chains tRNA → tRNA rRNA → Ribosomal subunits RNA uses the pyrimidine uracil instead of thymine; Uracil is only in RNA(with rare exceptions), Cytosine is in both RNA and DNA, Thymine is only in DNA(with rare exceptions) In RNA: Uses a Ribose sugar which contains a hydroxyl group attached to the second carbon In DNA: Uses 2­Deoxyribose, the second carbon is not bonded to a hydroxyl group. General features of RNA synthesis from a DNA template; mRNA mirrors the nontemplate strand of the DNA but uses Uracil in the place of Thymine. RNA is synthesized in a 5’ to 3’ direction like DNA. The DNA double helix is locally unwound during transcription; RNA polymerase Class notes: Every ribosome has multiple rRNAs that make it miRNA regulates translation, it is extremely small (micro hence mi prefix). Template strand read 3’ to 5’

Promoter in e.coli (prokaryotes) is ATATTA [­10 sequence] as well as AACTGT[­35 sequence]. E.coli RNA polymerase recognizes these and starts transcription. Local unwinding (transcription bubble) at Adenine and thymine rich location of sequence. AT takes less energy to unwind than GC due to less hydrogen bonds, therefore it is chosen since it is easier to unwind. In prokaryotes, genes are closely spaced and several can be transcribed on a single RNA molecule from a common promoter. Rho required for some termination sequences but the one in his example is rho independent; doesn’t need rho. As RNA is made it is folded into a hydrogen bonded hairpin loop. Intron not transcribed, they are spliced out. Eukaryotes have longer promoter. There is also a TATA box. RNA polymerase once bound to a gene will start transcribing in e.coli with no problem. Eukaryotic RNA polymerases (we have many) can be bound at a promoter but need help to start transcription. Thus promoter in eukaryotes helps to position polymerase but also recruits other proteins (general transcription factors) to get polymerase to start making RNA. It can be hard to determine the promoter in eukaryotes, can be thousands of base pairs. Need to do it experimentally at times. Poly(A)tail can be modified and adds stability to the material