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Exam 2 Study Guide

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Exam 2 Study Guide 2011

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These notes cover all of the concepts that Professor Dixon outlined in his slides. Please let me know if you have any questions or have suggestions as to how I can made these study guides more usab...
Biological Sciences II
Dr. Kevin Dixon
Study Guide
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This 10 page Study Guide was uploaded by rcg16b Notetaker on Saturday October 15, 2016. The Study Guide belongs to 2011 at Florida State University taught by Dr. Kevin Dixon in Fall 2016. Since its upload, it has received 179 views. For similar materials see Biological Sciences II in BSC at Florida State University.


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Date Created: 10/15/16
Lecture 10 Mendelian Genetics Reading: Chapter 14.1 Concepts • Spade foot toads- environmental factors can influence gene expression o Thyroxin levels can vary in individual tadpoles of the same species based on § Size of the pond § What they eat • Carnivores have bigger and shorter digestive tubes than omnivores o Traits are not dependent on the genotype, the trigger is the environment • Be able to explain the importance of inheritance to biology (and evolution in particular) o Helped discover chromosomes, DNA and ge nes o We learned that § we share about half of genes from mother and half from father § What trait is expressed in the child depends on the trait and the gen s • Be able to explain how Mendel did his experiments? o Mendel’s Model § Alleles - Alternative versions o f genes account for variations in inherited characters § For each characteristic, an organism inherits two copies (two alleles) of a gene, one from each parent § If the two alleles at a locus differ, then one, the dominant allele, determines the organism’s ap pearance o Cross pollinated two contrasting, true breeding pea varieties § True breeding- many generations of self pollination produced the same variety (homozygous) § self fertilized (used a paint brush to pollinate flowers) o Terms to know about his experiments : § Hybridization- mating of two true breeding varieties § P generation- true breeding parents (homozygous) § F1 generation- hybrid offspring (heterozygous) • All of his F1 progeny were monohybrids (heterozygous for one particular character being followed in the cross) § F2- when the F1 hybrids self pollinate (pundit square) o Discovered § ¾ were purple and ¼ were white in the F1 generation § the white allele is still lives! § *like in the toads, environmental factors can cause deviations in the 3:1 ratio § law of segregation- alleles separate during gamete formation and randomly reunite during fertilization § law of independent assortment- each pair of alleles segregates independently of each other pai r of alleles during gamete formation • Still confused? This basically means that if you have a parent that is AaBb, there is an equal change of passing on AB, Ab, aB, and ab allele pairs. • Be able to distinguish between Mendelian inheritance and blending inheritance? o Blending- offspring “blend” the traits of both parents (WRONG) § Would cause variation to go away completely o Mendelian- Mendel’s model, particulate inheritance § Each individual has two copies of genes and they each pass one on to offspring • Be able to explain the significance of the Law of Segregation o It is random what alleles are passed on o differences between the law of independent assortment and the law of segregation: § o The two alleles for a heritable character segregate (separate) during gamete formation a nd end up in different gametes randomly o An egg or a sperm gets only one of the two al leles that are present o Note: a parent passes on either their fathers allele or their mothers allele and environmental factors can affect gene expression and alter Mendel’s model up P p P PP Pp p Pp pp § 3:1 Phenotypic ratio p p P Pp Pp p pp pp § 1:1 Phenotypic ratio • Dihybrid cross o Previously we only considered monohybrid crosses § Example: pea color and pea shape 9:3:3:1 ratio YR Yr yR yr YR DD DD DD DD Yr DD DR DD DR yR DD DD RD RD yr DD DR RD RR § Y= yellow, y=green § R= round, r=wrinkled § Inside the chart: D=dominant, R=Recessive § 9 where both traits are dominant § 3 where Y is dominant and r is recessive § 3 where y is recessive and R is dominant § 1 where they are both recessive o Displays the law of independent assortment - segregation in one gene is independent of segregation in another gene § Segregation of one locus is independent of segregation at another locus § Mostly true, but not universal § A coin flip does not affect the other coin flip, black hair does not affect length of hair (50% chance of passing on an allele) • Terms: Gene, allele, locus, heterozygote, homozygote, genotype, phenotype, dominant, Recessive, F1, F2, monohybrid • Another example, you really only need the first column, You’re only going to get the A phenotype. 1:1:0:0 AB Ab aB ab Ab AABb AAbb AaBb Aabb Ab AABb AAbb AaBb Aabb Ab AABb AAbb AaBb Aabb Ab AABb AAbb AaBb Aabb Lecture 11 Reading - Chapter 13.1, 13.2 p 272- 276 • Be able to distinguish between the genetic consequences of sexual and asexual reproduction o Sexual § Two parents give a unique combination of genes to offspring § Not exact replicas o Asexual § Have exact genetic copies of themselves § Produces a clone- a group of genetically identical individuals § Genetic differences are due to mutations • Be able to distinguish the cellular processes behind asexual (mitosis)and sexual (meiosis) reproduction. o Mitosis o Meiosis • Know that chromosomes exist as homologous pairs in diploids o Chromosomes occur in pairs which implies that we have two copies of each gene • Be able to explain why meiosis produces haploid daughter cells from diploid parents • Know that sexual life cycles alternate meiosis and fertilization • Be able to explain how sexual life cycles vary in the timing of meiosis and fertilization. • Terms to Know - Haploid, diploid, homolgous chromosome (homologous pair), sister chromatid, gamete, zygote, meiosis, fertilization, gonad, sexual reproduction, asexual reproduction, clone, spore, gametophyte, sporophyte Lecture 13 - Meiosis Concepts • Watch this for clarification on meiosis • Gonad- testes and ovaries • Haploid vs Diploid o Diploid cell (2n)- pairs of homologous chromosomes, 2 sister chromatids at the start of meiosis/mitosis per chromosome (original amount of chromosomes) o Haploid cell (n)- when the cells divide the chromosomes in half through meiosis or mitosis o Don’t confuse sister chromatids and homologous chromosomes! § Homologous chromosomes - genetically similar but not necessarily identical § Sister chromatids of same chromosome- genetically identical from S phase • Be able to explain how meiosis produces haploid daughter cells by separating homologous pairs of chromosomes so that one homolog goes to each cell. o Sister chromatids are two copies of one chr omosome o after the chromosomes duplicate in interphase, the diploid cell divides twice o this yields four haploid daughter cells • Know that meiosis consists of two stages: meiosis I and meiosis II and know that the end product of meiosis is 4 haploid cells o Meiosis I § The diploid parent cell divides into two haploid cells with duplicated chromosomes § homologous chromosomes are separated producing two cells o Meiosis II § the haploid cells with duplicated chromosomes divide into haploid cells with unduplicated chromosomes. § sister chromatids are separated § basically the same as mitosis • Know the names of the stages of meiosis and what happens in each • Know that the pairing of homologs and crossing over both occur in prophase I and be able to explain the significance of both o Homologs- two chromosomes together= 4 sister chromatids o Homologs are aligned gene by gene o Allows for more genetic variation • Be able explain how the random assortment of homologs into daughter cells and the process of crossing over result in genetically variable gametes o Watch this!! It is less than 2 min long. -> o Cohesions hold together the sister chromatids o Synapsis- when the synaptonemal complex holds one homolog tight ly to the other § DNA breaks are closed up so that each broken end is joined to the corresponding nonsister chromatid § Therefore, a paternal chromatid is joined to a piece of maternal chromatid beyond the crossover point • Be able to explain the differences bet ween mitosis and meiosis o Mitosis § Genetically identical daughter cells § Development growth and asexual reproduction § Diploid to diploid or haploid to haploid o Meiosis § Daughter cells contain a subset of parent cell’s genes § Going from diploid to haploid cells § Haploid to diploid is fertilization § Each daughter cell has one member of each homologous pair § Necessary so that the next generation has the correct amount of chromosomes o 3 major differences § Synapsis and crossing over : • occurs in meiosis during prophase I while it doesn’t occur during mitosis § Homologous pair at the metaphase plate of meiosis: • chromosomes are positioned as pairs on homologs rather than individual chromosomes as in metaphase of mitosis § Separation of homologs: • at anaphase I of meiosis, th e duplicated chromosomes of each homologous pair move toward opposite poles but the sister chromatids remain attached. In anaphase of mitosis, sister chromatids separate. § Check this diagram out! -111220234407- phpapp01/95/biology-ch-13-11-728.jpg?cb=1324424834 (it is also in your book on page 261) • Vocabulary – Synapsis, crossing over, chiasma, meiosis I, meiosis II, segregation Readings Chapter 13.3, 13.4 Friday group assignment: • Law of Segregation • Law of Independent Assortment • Phenotypic Ratios Which of these three things is always true in all diploid, sexually reproducing systems? Law of Segregation What factors influence the other two? Law of Independent Assortment: location of genes relative to each other Phenotypic Ratios: depend on specific circumstances Lecture 14 Mendelian Inheritance of Two Traits Probability and Inheritance Reading: Chapter 14: 267 -271. • More on haploid/diploid o Haploid- one chromosome from each type o Diploid- two chromosomes from each type o The more homologous pairs, the more combinations (2 ) n=haploid number o n=23 for humans o if a diploid has 10 chromosomes, how many combinations are possible in gametes produced from this cell? § n=5 so 2 =32 • By able to explain why two characteristics can be inherited independently and describe the resulting pattern of inheritance. o synapsis is the joining of chromosomes in a homologous pair § prophase I results in crossing over § site of synapsis is called chiasma § DNA breaks and they exchange pieces § Makes 4 daughter cells, two of which have recombinant chromosomes § The chromosomes join at the exact same place to exchange the same loci • If they didn’t, one would have extra copies and one would be missing some genes § This varies genetic variation in gametes § Allows loci to segregate independently § AABb 2n=4 • A and B are on two separate chromosomes A A A A B B b b • Be able to explain how mei osis relates to the Law of Independent Assortment o first division- each pair of homologs sort its maternal and paternal homologs into daughter cells o happens independently of every other pair o Each daughter cell now is one possible outcome of all combinations of maternal and paternal chromosomes o The diploid parent cell divides into two haploid cells with duplicated chromosomes • Be able to apply the sum rule and product rule of Probability correctly o Sum probability § of either one of two different probabilities occurring = P (1) + P (2) § If the word ‘or’ occurs in a statement about probabilities, then the probabilities are added § Example from class: 1/8 + 1/8 + 1/8 = 3/8 (GGB, BGG, GBG) o Product Probability § two independent events happening = P (event 1) X P (event 2) § If a statement about the probability contains an ‘and’ the n the probabilities are multiplied § Peas • Be able to apply the rules of probability to genetics problems o P(Pp) = P(Pfirpseco) + P(Psecopfir) = 0.5*0.5 + 0.5*0.5 = 0.25 + 0.25 = 0.5 o AaBb= four types of gametes § How do we calculate the probability? § Half gametes will have (A) half have (a) § Half have B and half have b § AB=Ab=aB=ab= 0.5*0.5=0.25 o Probabilities of the F2 genotypes § AaBB § P(Aa)*P(BB)=0.5*0.25=0.125 o Cross AAbbCcDdEe x aaBbCcDDee = AabbccDDee = 1*0.5*0.25*0.5*0.25= 0.0625 Lecture 15 Reading: 15.1, • Know that humans are more tolerant of sex chromosome variation than is true for autosomes o XY System- sex of an offspring depends on whether the sperm cell contains an X chromosome or a Y o X0 System- in some insects, only one type of sex chromos ome, females are XX and males are X0 § Sex is determined by whether a sperm cell contains an X chromosome or no sex chromosome o ZW System- birds, fishes and some insects, the sex chromosomes present in the egg determines the sex of the offspring. § Females are ZW and males are ZZ. o Haplo-diploid System- no sex chromosomes in most bees and ants. § Females develop from fertilized eggs (makes them diploid). § Males develop from unfertilized eggs (makes them haploid with no father). § Have more control about the sex of t heir offspring • Be able to list and explain the different types of sex determining mechanisms o Genes location on sex chromosomes is really the main thing • Know that the SRY region determines sex in humans o X chromosome is large o Y is smaller and all for male things o Gene on the Y chromosome required for the development of testes o Sex determining region of Y o Without SRY, the gonads develop into ovaries, SRY region makes males males o SRY codes proteins that regulate other genes o Having an extra chromos ome is much worse in mammals • Understand that sex determination affects the expression of many genes o Sex-linked gene- gene located on either sex chromosome § X-linked genes § Y-linked genes o Turner’s Syndrome X § Short stature, sterility, stocky build, little brea st development, swollen hands and feet, risk of cardiac problems, thyroid problems, spatial and mathematical cognition § Having extra of sex chromosomes is not as big of a deal o Klinefelter’s Syndrome XXY § Effective sterility, youthful appearance, rarely incre ased breast tissue, language learning disability (usually little effect in adult males other than sterility) o XXX § Basically indistinguishable from XX females § Some menstrual irregularities, taller than average, may have mild learning disabilities. o XYY § Minor differences from XY males § More rapid growth, taller than average, likely have mild learning disabilities. o Most XXX and XYY individuals are never diagnosed o Temperature dependent sex determination § Species that have this do not have sex chromosomes § Common in turtle and the only kind in crocodiles • Terminology - XY, ZW, XO, haplodiploidy, temperature -dependent sex determination, SRY Lecture 15 part 2 – sex-linked inheritance Reading: 15.2 • Be able to explain how loci on the X chromosome are inherit ed differently in males and females. o X - large, many genes o Y - small, few genes mostly associated with ‘maleness’ § White eye associated with male and red with female in flies § F2 produced a 3:1 ratio, but all of the red ones were female and the white was al ways male § White eye locus is on the X chromosome § Males pass on wildtype to daughters and Y to their sons • Be able to recognize the pattern of X -linked inheritance • Know that in female mammals one X chromosome is inactivated in each cell. o X inactivation results in heterozygotes that are chimeras Lecture 16: Linkage Continued and More Complex Human Genetics • Chimeras- different phenotypes in different cells o Orange allele overrides any colored loci (epistasis) o Orange males more common because they on ly need one copy of the orange o Heterozygotes- patches of different colors of fur, called tortes shell o Black male crossed with orange female X Y X X X X Y X X X X Y § ½ will be orange males § ½ will be non orange females • Know that distance between loci on a chromosome affects the probability of crossing over occurring in between the loci o Greater distance between two genes= more points where crossing over can occur • Be able to explain why loci close together on a chromosome are likely to be inherited as a unit and are said to be linked. o Genetic recombination - the product of offspring with combinations of traits that differ from those found in either P generation parent o Crossing over accounts for this • Be able to explain the difference between parental a nd recombinant genotypes and phenotypes. o Parental types- offspring that look like parents - chromosomes haven’t changed § Closer and no crossing over o Recombinant types- do not look like parents - chromosomes have changed and crossed over § Farther apart with crossing over • Know that linkage is important in mapping genes and in hunting for loci with effects on a particular phenotype. o Linkage- loci on the same chromosome are linked o Genetic map- an ordered list of genetic loci along a particular chromosome o Hypothesis: recombination frequency depends on the distance between genes on a chromosome § The farther apart the two genes are, the higher the probability that a crossover will occur between them= higher recombination frequency § To study this, he mapped genes o Linkage map- genetic map based on recombinant frequencies § Unlinked- recombinant and parental genotypes in gametes are equally likely (independent assortment) § Linked- parental genotypes are more likely than recombinant • If completely linked, you will get 3:1 rat io • If linkage is not complete, some other ratio than 9:3:3:1 or 3:1 • Do not do dihybrid cross to see if there is a link • Do a test cross- take F1 and cross it with a homozygous recessive individual o Example § Black (b) and vestigial (vg) in Drosophila (both r ecessive) § vg vg b b x vg+ vg+ b+ b+ à vg+ vg b+ b § vg+ vg b+ b x vg vg b b à 1:1:1:1 if independent assortment vg+ b+ vg b vg b+ vg+ b vg b § if linkage, recombinants are more common (the last two). The closer vg and b are to each other, the less likely it is to happen. § If linkage: vg+ b+ vg b vg b+ vg+ b vg b >1/4 >1/4 <1/4 <1/4 o equal outcomes would indicate that there is not linkage o should never have significantly more recombinant than parental • Know that genetically influenced diseases can a lso be influenced by the environment. o Different genotypes react differently to environment • Males are more likely to show a recessive trait - only have one x o Can show up in females but have to be homozygous o Males get the more common trait from their mother o White eyed male crossed with wild type female § ¼ will be wild type males X Y w+ w+ w w+ X X X Y X X X X Y X Lecture 17 Gene Function and Dominance • Be able to explain that incomplete dominance means heterozygotes have an in termediate phenotype o both alleles are expressed to give an intermediate form between homozygotes o looks like blending but is not because ¼ of them are the dominant allele, ¼ are the recessive allele and ½ have incomplete dominance. o Two copies makes red, on e copy makes pink and no copies makes white o The reason there are differences in relationships between alleles is because of what the genes are doing o Albino has a change in the coding sequence making a different amino acid chain • Be able to explain that cod ominance means that heterozygotes have a different phenotype than either homozygote o both alleles are expressed to give a different phenotype than either homozygote o expressing both alleles, but separatelyd • Be able to solve problems using the inheritance of ABO blood types o ABO blood groups are determined by three alleles: I , I , and i o A person’s blood group is either A, B, AB, or O o A and B are two carbohydrates § have carbohydrate A (type A blood) I I OR I i § have carbohydrate B (type B) I I OR I i § have both (type AB) I I § have neither (type O) ii o How does ABO Inheritance Work? o I I X I I = A or AB A B o ii X I I = either A or B o I i X ii = B or O • Be able to give examples of how gene function is involved in the type of dominance • Be able to trace the inheritance of traits on a pedigree and interpret the the type of inheritance from the pedigree o Pedigree- Traits of parents and children across generations in a family tree o Pedigree and genetic markers - looks for makers that are associated with the disease o What is the pattern with a dominant trait? § Widow’s peak o Recessive trait? § Earlobe • Know the inheritance of Huntington’s Disease, Cystic Fibrosis and Sickle Cell Anemia. o Cystic fibrosis- the most common lethal disease in the US § Normal allele for this gene codes for a membrane protein that transports chloride ions between cells and the extracellular fluid • chloride transport channels don’t work in the plasma membranes there are two recessive alleles for cystic fibrosis § causes high concentration of extracel lular chloride = extra thick, sticky mucous • builds up in the pancreas, lungs, digestive tract, and other organs o Sickle Cell Anemia § Most common inherited disorder of Africans § caused by the substitution of a single amino acid in the hemoglobin protein of red blood cells § if homozygous, all hemoglobin is sickle-cell § When the oxygen content of blood is low, the sickle -cell hemoglobin proteins aggregate into long fibers that deform the red cells into a sickle shape § Sickled cells may clump and clog small blood vessels § Dominant/recessive depends on what phenotype you look at o Huntington’s disease § a degenerative disease of the nervous system § caused by a lethal dominant allele that has no obvious phenotypic effect until later in life § irreversible and fatal § a child born to a parent with the Huntington’s disease allele has a 50% chance of inheriting the allele and the disorder § it is dominant which is unusual Reading 14.3 (271-273 (end of multiple alleles), 14.4, 17.1 and 17.4 (review only)


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