Genetics Test 1 Study Guide
Genetics Test 1 Study Guide Bisc 336
Popular in Genetics
verified elite notetaker
Popular in Biology
This 19 page Study Guide was uploaded by Anna Ballard on Sunday September 11, 2016. The Study Guide belongs to Bisc 336 at University of Mississippi taught by Ryan Garrick in Fall 2016. Since its upload, it has received 107 views. For similar materials see Genetics in Biology at University of Mississippi.
Reviews for Genetics Test 1 Study Guide
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 09/11/16
LECTURE 1 Recap on Vocab: ________ – the study of heredity and variation Heredity – ___________________________________ Variation – differences between individuals Understand which major discoveries underpin recent developments in genetics 1600-1850 _______ – (epigenesis) egg differentiates into adult structures over time - Aka development Pasteur – (spontaneous generation) living organisms do not derive from non- living components _______ – (evolution) _______ ____ ____________, with _______ _________ as the driver of change - Variation in a population - Competition for _______ - Advantages over others 1600-1850 ________ and “unit factors” – predictable patterns of inheritance: random segregation, assortment • Chromosomal theory – _____________________________________________________ - process of ________ and _______ • Transmission Genetics – _____________________________________________ - chromosome and pedigree analysis Q: Who determined that DNA was the carrier of genetic info? What was believed previously? A: Avery at Al. – People believed previously that proteins were the carriers of genetic info Q: What about Watson and Crick? A: Watson and Crick discovered that DNA was organized as a double helix with a phosphate backbone and nucleotide pairing rules (___ goes with T, and ___ goes with G) Describe relationships between components of the central dogma; contrast DNA, RNA, and proteins. THE CENTRAL DOGMA DNA ______________ ____________ ________________ Translation (Matching Vocab) A. Allele B. Genotype C. Gene D. Genome E. Phenotype F. G. 1. __________ how a person’s genotype interacts with environmental factors (observable/physical characteristics) H. 2. __________ section of DNA that codes for RNA to make proteins, which in turn make us I. 3. __________ an entire haploid set of chromosomes aka what it takes to build an entire organism (“blueprint”) all of the chromosomes in the nucleus of a cell J. 4. __________ simple combination of the alleles at a locus K. 5. __________ different forms or variants of a gene L. M. PROTEINS N. Definition: the end products of ________ _____________ O. P. Remember: All _______________ are proteins but not all proteins are _______________! Q. R. Q: What is Sickle Cell Anemia? S. A: Sickle Cell Anemia is a disorder that arises from a mutant gene which produces a mutant protein that alters phenotype. The mutation is __________________, which is a single nucleotide change that leads to a different amino acid, and a ß-globin protein T. U. Recall the series of advances that were possible following the development of recombinant DNA. V. RECOMBINANT DNA W. Q: Explain DNA cloning. X. A: _________________ cut DNA at specific nucleotide sites, then ___________ (plasmids) take up the DNA fragments from the restriction enzymes. After that, ___________________, such as bacteria, take up the vector. As the bacteria colony grows, clones are made. Y. Z. Q: What are “Model Organisms” and their characteristics? AA. A: Model organisms are a diverse set of organisms that can be well understood from a scientific standpoint based on these characteristics: 1. ______________________ AB. 2. _______________________________ AC. 3. ______________________________________ AD. AE. BIOTECHNOLOGY AF. AG. ___________________________ - crop plant resistance (to insect herbivores, viruses, etc), nutrient enhancement, milk production, etc.. - controversy: o engineered traits might “escape” into wild relatives o patents on genes/living organisms is not OK AH. AI. ___________________________ - rapid screening of disease-associated alleles - determine risk of developing (or having a child with) a genetic disorder - develop individual-based therapy, based on observed expression of genes (e.g., in cancer cells). AJ. AK. NEW FIELDS AL. A. Genomics B. Proteomics C. Bioinformatics AM. AN. 1. __________ identify proteins present in a cell at a given time/location; modifications and interactions AO. 2. __________ filtering and analyses of very large genetic datasets; software development AP.3. __________ structure and function of genomes (all of the DNA carried in an organism), often comparative AQ. AR. LECTURE 2 AS. Know the terminology (cells, chromosomes, etc.) AT. Cell Structure and Genetic Function A. Nucleus B. Organelles C. Cytoplasm D. Cytoskeleton AU. AV.1. __________ colloidal matrix AW. 2. __________ mitochondria, ER, centriole, chloroplast AX. 3. __________ houses DNA AY.4. __________ lattice of tubules and filaments AZ. BA. In the Nucleus A. Chromatin B. Chromosomes C. Nucleolus BB. BC. 1. ___________ where ribosomal RNA (rRNA) is made and where ribosomal subunits are initially built BD. 2. ___________ loosely or uncoiled DNA that is unattached from histone; present when cells are NOT dividing BE. 3. ___________ condensed (super coiled) DNA with discrete structures, able to be seen just before and during mitosis and meiosis, otherwise they are more loosely packed BF. BG. In the Cytoplasm A. Smooth Endoplasmic Reticulum (SER) B. Rough Endoplasmic Reticulum (RER) C. Mitochondria D. Chloroplasts E. Centrioles BH. BI.1. __________ maternally inherited organelle in charge of oxidative respiration (make ATP); its chromosomes live outside of the nucleus (no heterozygous or homozygous copies of genes because they are haploid/diploid; “Powerhouse of cell” BJ.2. __________ sites of biochemical reactions such as fatty acid and phospholipid synthesis BK. 3. __________ involved in photosynthesis in all plants, algae, and some protozoa (but not all cells) BL. 4. __________ site of biochemical reactions such as translation and covered in ribosomes BM. 5. __________ used in mitosis and meiosis – in charge of the organization of spindle fibers which attach to and move chromosomes BN. BO. Explaining Chromsomes BP.• When present as sister chromatids (2 chromatids), chromosomes have a _____________ used for __________ BQ. • _________ – DNA that may or may not be transcribed or translated BR. BS. BT. BU. BV. BW. BX. BY. BZ. CA. CB. CC. CD. CE. CF.Homologous Chromosomes CG. • All members of a species have the same number of ____________ per somatic cell CH. • Occur in pairs, 1 from the maternal and 1 from the paternal parent; matching _______, ____________, and __________________ CI.• Sex chromosomes (although dissimilar) still behave as ___________. CJ. CK. Karyotype CL. • Humans: - Diploid (2n) # = _________ - Haploid (n) # = __________ CM. CN. Understand the function of mitosis CO. Mitosis CP.• genetic continuity between generations of cells CQ. • Parent and daughter cells have __________________ CR. Function of mitosis: replacing skin and hair cells; growing CS. CT. Describe steps involved in mitosis CU. CV. Regular Function: CW. • ________ – cells performing normal function (ex: blood cells carrying O2) CX. • ________ – non-dividing cells withdrawn from the cycle heading towards duplication; continue carrying out normal function CY.• ________ – where DNA replication takes place CZ. • ________ – similar to G1; cell is still performing normally but form of chromosomes has been altered. DA. DB. Mitotic Division (PMAT) DC. Q: What happens in Mitosis, Prophase? DD. A: ___________ move to pole and organize microtubules and spindle fibers, _________________ breaks, and ________________ are visible as pairs of genetically identical sister chromatids DE. DF. Q: What happens in Mitosis, Metaphase? DG. A: Chromosomes migrate to the central axis and then align along the metaphase plate - Sister chromatids pulled apart in the ___________ direction DH. DI.Q: What happens in Mitosis, Anaphase? DJ.A: Sister chromatids separate and migrate to opposite poles, separated sisters = ______________. Daughter cells will get ___________ sets of chromosomes DK. DL. Q: What happens in Mitosis, Telophase? DM. A: Cytoplasm starts to divide (cytokinesis), chromosomes uncoil back to diffuse chromatin, __________________ reforms, and _________________ disappear. DN. Order of Regular function and Mitosis: DO. Interphase 1. _________ 2. G0 3. _________ 4. G2 DP. Mitosis 1. Prophase 2. ___________ 3. ___________ 4. ___________ DQ. DR. LECTURE 3 DS. Mitosis Recap 1. Starts with _____________ (2n =4) 2. __________ – main events: chromosomes condense, nuclear envelope degenerates, chromosomes move to opposite sides and lay down spindle fibers 3. __________ – 4 chromosomes each consisting of a pair of sister chromatids line up in middle 4. __________ and __________ – creates 2 new diploid (___) daughter cells DT. DU. Describe Steps involved in Meiosis DV. DW. Function of Meiosis: DX. • Continuity between generations of cells DY.• Reduces number of chromosomes by ½: diploid –> haploid DZ. • Passes genetic material from parent to offspring EA. • Fusing ½ of each parent DNA allows them to reform to create the next generation’s makeup EB. EC. • Meiotic cell cycle involves ONE ______________________ and TWO _________________ - PMAT I and PMAT II ED. A. Prophase I B. Metaphase I C. Anaphase I D. Telophase I E. Prophase II F. Metaphase II G. Anaphase II H. Telophase II EE. EF. EG. 1. ___________ – Chromosomes mostly visible, homologous chromosomes pair up (synapse) together, ________________ from physical contact between chromatid of one chromosome with the chromatid of another chromosome allows for crossing over - synapsed homologs have how many chromatids? ________ 2. ___________ – chromosomes migrate and align in middle; equational division starts 3. ___________ – tetrads pulled apart (one pair of _____________ to each pole), poles now have random mix of maternal and paternal sister chromatids that have been previously modified by crossing over 4. ___________ – starts with 2 separate daughter cells from meiosis I; chromosomes condense, centrioles move to opposite poles, spindle fibers laid down; no pairing of homologous chromosomes so no ____________; chromosomes in each cell have already been duplicated (are now a pair of sister chromatids) 5. ___________ – 1 division (___________) ends here; nuclear membrane forms around chromosome sets at each pole, cytoplasm begins to pinch; interphase occurs briefly after without DNA replication 6. ___________ – chromosomes migrate and align in middle; reductional division starts 7. ___________ – 2nddivision (___________) ends here; cytoplasm is divided, nuclear membrane reforms; ultimately 4 haploid cells (gametes), each with different genetic makeup, are produced from single meiotic cycle 8. ___________ – dyads pulled apart (one __________ to each pole); poles now have random mix of chromatids from each chromosome; each full (hapoloid) set of the genome EH. EI.Contrast Mitosis vs. Meiosis EJ. ** See class handout ** EK. EL. LECTURE 4 EM. Contrast spermatogenesis v. oogenesis EN. Spermatogenesis EO. • General procedure of meiosis EP. EQ. Male Gamete Production ER. • Starts with ______________________________ (spermatogonium) ES. • Becomes an enlarged primary spermatocyte, which undergoes 1 (______________) meiotic division ET.• _____________________ undergoes 2 ndmeiotic division EU. • each make 2 haploid spermatids that develop into motile sperm = total of 4 _________ sperm EV. EW. Oogenesis EX. Female Gamete Production EY.• Starts from ____________, which enlarges to become the primary oocyte - primary oocyte then splits during ____________ EZ. FA.• Daughter cells of the primary oocyte receive equal _________________, but unequal _____________ (cellular volume) st nd - the cell that gets the least is called either the 1 or 2 _______________ and does not continue through the process of oogenesis FB. • The cell that gets more cytoplasm is either called the _____________ (after 1 division) or ______________ (after 2ddivision) FC. • Oocytes that get shorted cytoplasm do not become eggs FD. • Just ___________________ (egg) produced per meiotic cycle FE. FF. Sperm v. Egg Production FG. • Initially start off similar – diploid parent cell undergoes growth and maturation FH. • End of first division (reductional): - first polar body in oogenesis - 2 daughter cells as secondary spermatocytes in spermatogenesis FI. • A single ovum produced at end of oogenesis vs. 4 haploid sperm at end of spermatogenesis - haploid gametes produced are genetically different from parent cell FJ. FK. Spontaneous DNA Mutations FL.Usually caused by rare erros during DNA replication, at low, on-going “background rates” like a slow ticking clock FM. • Most mutations are _________ (such as in non-coding DNA so there is no phenotypic effect) FN. • Most mutations in coding regions are detrimental, only some are beneficial - most common outcome: production of an allelic variant that does not function as well as original DNA FO. FP.Q: What are the different Mutations in Protein-Coding genes? FQ. A: __________ – change in a codon and in its amino acid –> changes amino acid/protein and ____________ – change in a codon that creates a premature STOP codon so protein synthesis is terminated FR. FS. Q: What could create new allelic variants that did not previously exist? FT.A: Crossing over between parts of homologous chromosomes in ______________ because it creates “mosaic-like” chromsomes FU. FV.Q: What is independent (random) assortment? FW. A: This is the ultimate outcome of meiosis because each haploid gamete is not a clone of its diploid parent. FX. FY. Q: Why is it important to know about sickle-cell anemia when learning about genetics? FZ. A: Sickle cell anemia is caused by a single DNA substitution, which causes a mutant protein to be produced, which in turn alters the phenotype of an individual, effecting his or her red blood cells. GA. GB. Q: What is the difference between a population and a gene pool GC. A: ______________ – group of individuals living together that have capacity to interbreed and can also compete for limited resources; ______________ – total number of allelic variants of a gene that exist within a particular group of individuals GD. GE. LECTURE 5 GF. Understand Mendel’s 4 Postulates GG. Q: Who was Gregor Mendel? GH. A: A monk who experimented with pea plants while working in a monastery garden – discovered units of inheritance by observing similarities in ______________ between one generation to the next without any knowledge of _________________________ GI. GJ.Q: What traits did he observe were encoded in the same gene? GK. A: flower color, position of flower, seed color, seed shape, pod shape, pod color, and stem length. Pea plants are a great model organism because they are easy to cross, fast growing, and have observable traits GL. GM. Q: What does “true breeding” mean? GN. A: This means that the individuals were homozygous for a gene GO. GP. Monohybrid Cross GQ. • Mating between individuals that differ in ________ trait GR. • P1 – parentals – true breeding green or true breeding yellow (RR ast rr) - F1 – stgeneration offspring –> ______ % yellow (all Rr) - F2 – 1 generation crossed with each other –> ______ % yellow, _______ % green (RR, rr, and Rr) o _________ ratio o – things that pass unchanged from one generation to the next, and determine the phenotype that is expressed GS. GT. Mendel’s First 3 postulates GU. 1. ________ of a gene: occur in pairs in an individual; only applies to diploid organisms (including peas) GV. 2. ________________ : one phenotype dominates over the other (not true of all genes) GW. 3. ________________ : during meiosis, chromosomes on which alleles reside separate and go to different gametes randomly GX. GY. Use a Punnett Square for Mono- and Di-hybrid crosses GZ. HA. Q: If a cross between P1 resulted in an F1 generation of all heterozygotes, what will the F2 generation look like genotypically and phenotypically? HB. A: genotypic ratio = 1:2:1 and phenotypic ratio = 3:1 (See Punnett Squares Below) HC. HD. HE. Punnett Squares Practice A.F1 HF. HG. Cr. P1 B.D C.d D osD Cro s ss E.__ F. __ HJ.____D.DK. HId _ _____ _ _ H.__ I. __ G. d _ _ HL. HN. HM. _____ ___ d HO. HP. Dihybrid Cross • Simultaneously follow inheritance of 2 phenotypic characters • Generates an expected phenotypic ratio 9:3:3:1 HQ. HR. yellow, round X green, wrinkled HS. HT. HU. HV. HW. F1 HX. All yellow, round HY. HZ. IA. IB. F1 X F1 IC. yellow, round X yellow round ID. IE. IF. IG. IH. 9/16 yellow, round 3/16 green, round II.3/16 yellow, wrinkled 1/16 green, wrinkled IJ. IK. IL. • As in monohybrid cross, more genotypes than phenotypes • Here, 9 genotypes underlie the 4 phenotypes GGWW, GGWw, GgWW, GgWw, GGww, Ggww, ggWW, ggWw, ggww • 9:3:3:1 ratio requires that 2 traits are fully independent • fig 37 IM. IN. IO. IP. Extend Principles of Independent Asortment into three-trait crosses, suing the forked line method IQ. Forked Line Method (Phenotypes) IR. IS. IT. round (3/4) yellow, round IU. 3/4 X 3/4 = 0.57 (9/16) IV. yellow IW.¾ IX. wrinkled (1/4) yellow, wrinkled IY. 3/4 X 1/4 = 0.19 (3/16) IZ. JA. JB. JC. round (3/4) green, round JD. 1/4 X 3/4 = 0.19 (3/16) JE.green JF. 1/4 JG. wrinkled (1/4) green, wrinkled JH. 1/4 X 1/4 = 0.06 (1/16) JI. JJ. JK. LECTURE 6 JL. Mendel’s 4th Postulate JM.• ______________: different phenotype traits are inherited independently of one another JN.• The 2 alleles of one gene have an equal chance of segregating with either 2 alleles at another gene JO.• Need a min. of 2 different genes, each with 2 alleles JP. JQ.What about a trihybrid cross? JR.• identical processes of segregation and independent assortment apply to 3 (or more) traits JS.• Punnet Sqare with 64 boxes (crikey!), or forked line method (http://www.youtube.com/watch?v=nBTX2h2GYYw) JT. • Based on laws of probability and relationships between expected phenotypic ratios JU. JV.Expectations v. Observations JW. • We can calculate expected phenotypic ratios from Mendel’s principals (segregation, independent assortment) JX.• …and we can make observations of phenotypic ratios from experimental crosses JY.• Next, we need a way to assess the “fit” between observed v. expected ratios (i.e, equivalent or not?) JZ. KA. Recap: Mendel’s Postulates KB. 1. Alleles of a gene occur in __________. KC. 2. _________/__________ : one phenotype dominates KD. 3. _________________: during meiosis, alleles of a gene separate and go to different gametes randomly KE. 4. ____________________ : different phenotypic traits are inherited independently of one another KF.- rule or inference only possible following experiments that follow the inheritance of 2 different phenotypes separately such as pod color and pod shape KG. KH. • If we want to assess whether Mendelian rules of inheritance are true, we need a sufficient number of observations to really assess that - why larger sample sizes converge on true answer better than small sample sizes KI. KJ. Perform Chi-Squared Analyses, Determine P-values, and interpret them with respect to a null hypothesis KK. Testing Mendel’s Principles KL. • Assuming dominant/recessive, we can test for ___________________ and _______________ _____________________. KM. • A dihybrid cross needs a larger sample size than monohybrid because there are more combinations of phenotypes KN. KO. Q: What is your null hypothesis when performing chi-square analyses KP.A: The null hypothesis would be whatever ratios you would get under Mendelian principles (Monohybrid: 3:1 genotype; Dihybrid: 9:3:3:1 genotype) - you either ____________ your null hypothesis or _______ ____ __________ (never accept a null hypothesis for Mendelian X^2 Analyses) KQ. KR. Q: What is the Chi-Square equation? KS. A: X^2 = ∑((o – e)^2 / e) where ____ represents observed and ______ represents expected KT. KU. Q: How do you find your degrees of freedom? KV. A: Count the number of classes (ex: round, wrinkled, etc.) and subtract by 1 (n – 1) KW. KX. Q: How do you use a chi square look up table? KY.A: After you determine your chi-square number and your degrees of freedom number, look at the given table (FIG 3-10) and find which row represents your degrees of freedom. If your df=1, look on the first row, etc. Next, find where your X^2 value would fall on that row. If the P value above that is _____________ 5%, or 0.05, reject the null hypothesis because this means that there is less than 5% chance of obtaining observed data when the null hypothesis is true (aka we start to think that there may be other factors such as crossing over and independent assortment messing up the data) KZ. LA. Q: You mean to accept the null hypothesis when your p value is > 5% (p > 0.05) LB. A: NO!! This is when you would “Fail to reject” the null hypothesis LC. LD.Using a X^2 test “look up graph” LE.• Chi square of 0.53 •locate area on X axis where 0.53 occurs LF.• Df = 1 •diagonal lines represent degrees of freedom values LG.• Find 0.53 on xaxis, draw line up to df 1 diagonal line LH.• Cut across to y axis; provides estimate of pvalue LI. FIG 310 LJ. LK.Reasons for Rejecting a Null Hypothesis (P<0.05) LL. • One or more of the underlying assumptions do not hold: LM. 1)dominance/recessiveness, 2) segregation, and 3) independent assortment LN. LO. LP. LECTURE 7 LQ. Use Pedigrees to Infer Mode of Inheritance of a Trait LR. Pedigrees LS. • __________________ – is the phenotype of interest dominant or recessive? Is it an autosomal or X-linked chromosome? - Recorded histories of family trees - Circles are females, squares are males, diamond for unkown sex - Shade shape if individual is expressing a phenotype (ex: albino) - Parents are unrelated – ________ line - Related parents – _________ line - Numbers – represent _______________ LT. LU. Pedigree Analysis LV.• Key point – ususally does not provide same level of certainty as carefully designed _______________________ LW. • Larger pedigrees allow _________ inferences to be made by providing greater confidence in __________ of ______________. LX. • Multiple pedigrees _______________________. LY. LZ. Modification of Mendelian Ratios Ch. 4 MA. After Mendel: MB. Research started to focus on traits that do not follow the simple Mendelian model MC. • Although the basic mechanism of inheritance is the same, other assumptions did not always hold - Some traits are influenced by more than one gene - Can also be influenced by the environment - Maybe dominant/recessive doesn’t have same effect such as when occurring in the MD. heterozygous form ME. MF. • _______________ – the (often dominant) allele carried by the most common phenotype in a population (common and usually dominant) MG. • ___________ is the source of new genetic variants (mutant alleles) and are more rare MH. MI. MJ.More Genotype Notation MK. • Capital letter alleles don’t have to represent dominance only; they could also represent a ________________ allele (e.g. as D+) ML. • Also don’t always have to represent with one alleles (e.g., the Wr allele) which can also be separated by a slash (e.g. Wr+/Wg MM. MN. 1. Single-Locus Phenomena MO. - single locus/gene and one gene is controlling a specific trait MP. - departures from Mendelian genotypic ratios MQ. MR. Incomplete Dominance MS. • Offspring’s phenotype is an ______________________ of the parents (think calico cats) MT. • In a monohybrid cross, each of 3 genotypes (homozygous dominant, heterozygous, homozygous recessive) yields a ___________________________. MU. • Expected phenotypic and genotypic ratios = ____________ MV. MW. Co-Dominance MX. • Offspring’s phenotype is different form the 2 parents, ___________________ - Joint expression of both alleles in ___________ - Expected genotypic and phenotypic ratios in monohybrid cross = 1:2:1 MY. MZ. Multiple Alleles NA. • Strictly a population-level phenomenon NB. • Ex: Human ABO blood group – controlled by a single autosomal locus with 3 alleles (I^A, I^B, and I^O) - 3 alleles = ________ possible diploid genotypes NC. ND. • These 6 genotypes yield 4 alternative phenoytpes - A&B is _______ dominant to O but _______________ to each other 1 AO or AA – _______ is expressed 2 AB – both expressed 3 BO or BB – _______ is expressed 4 OO – neither A nor B expressed NE. NF.Recessive Lethal Alleles NG. • Selective disadvantage (or advantage) of a given allele causes deviation from Mendelian ratios NH. • Some ___________________ can be tolerated by ______________, but ot _____________ NI. • ____________________: individuals that are homozygous for the recessive allele have a short life expectancy •sometimes a single copy of the recessive allele is enough to cause problems NJ. NK. Agouti has complex inheritance NL.• This locus affects two traits (hair color and embryonic development); dominance/recessiveness switches • The yellow allele (AY) is a lethal recessive (AY/AY is never seen – embryos die before birth) 1 homozygotes with that genotype die very young 2 • AY is dominant with respect to coat color (AY/A survives but has an unusual yellow coat) • The dual action of this locus on survival and coat color cause __________________ from Mendelian ratios 1 we know from a series of crosses that if we cross a pair of agouti mice (AA X AA), all offspring will show the agouti color, and therefore they all survive. 2 NM. yellow (AAY) X yellow (AAY) = _____ yellow and _____ agouti, the survivors (the other 1/3 die due to detrimental homozygote genotype so you do not account for these in ratios) NN. NO. agouti (AA) X yellow (AAY) = ½ agouti and ½ yellow (all survive) NP. NQ. NR. Dominant Lethal Alleles NS. • ____________________: degeneration of nerve cells in the brain, caused by a run of “CAG” repeat, chromosome 4. NT.• Usually late onset (around 40 years) in heterozygotes NU. • Alleles can persist in a population when carrier individuals survive long enough to reproduce. NV. NW. Q: So, Why has Huntington’s diseas not been weeded out of the genepool? NX. A: because of those who are heterozygous for the trait have a later onset and can mate before showing signs or symptoms. NY. NZ.Genes on Sex Chromosomes OA. • __________: compared to autosomal genes, those on the Xchromosome have unique inheritance OB. • ___________________ only has ONE copy (e.g. XY males, in mammals) OC. OD. Xlinkage and Drosophila White eyes OE. OF. SexLimited Traits OG. • Autosomal genes that act differently in sexes OH. • e.g., feathers of male and female chickens, where a certain phenotype found only in one of the sexes • Hormones determine expression OI. OJ.SexInfluenced Traits OK. • Autosomal genes that act differently in sexes, but the phenotype is not limited to only one of the sexes J. M. P. OL.ex: baldness K. N. Q. OM. L. G O. F R. M ON. e e a OO. n m l OP. OQ. ot a e OR. y l p e OS. e OT. OU. S. U. W. OV. TBB V. B X. B a a OW. l l OX. d d OY. OZ. Y. AA. AC. PA. Z. B AB. b Not AD. PB. Bald PC. PD. • Mitochondria and Chloroplasts have their AE. AG. AI. own proteincoding genes, so can affect AJ.N phenotype AF.b AH. o b Not t PE.• Since they are haploid and uniparentally inherited, ___________ follow basic Mendelian Principles PF. PG. LECTURE 8 PH. 2. Locus-by-locus interactions PI. PJ. Physical Linkage - A form of ___________________________ - Two genes that occur closely on the same chromosome will have alleles hthat exhibit ____________ because they tend to be inherited as a single unit - Probability of crossing over to separate the 2 is possible but not very likely PK. PL.Locus-By-Locus Interactions PM. • Many phenotypic traits are controlled by more than one gene PN. • interactions among genes and their products can be direct: - In some cases, alleles at one locus __________ the expression of alleles at another (epistasis) - __________________ are masked by an epistatic locus PO. PP.Epistasis: The Bombay Phenotype PQ. • The single locus that determines human ABO blood type can be impacted by a second locus, ___________ - the recessive allele at ____________ causes individuals that are AA, BB, or AB to express the O blood phenotype - ABO blood group are the _________________ alleles and that locus is the ___________ locus PR. Ch. 5 –> Sex Determination and Sex Chromosomes PS. PT. Sexual Differentiation • sex chromosomes are the location of many (but not all) sex determining genes • mechanisms of sex-determination are _______________ among eukaryotes (not just ‘XY’ vs. ‘XX’) • differences between males (♂) and females (♀) (sexual dimorphism) are separated into 2 classes: PU. 1. ___________: gonads only PV. 2. _____________: other traits related to gender PW. PX. Homo- vs. Heterogametic • the ________________ (XY) has a pair of dissimilar sex chromosomes (vs. ___________________(XX)) • human males make _____________________ (some with X, others with Y) that determine sex of offspring PY. • ______________________: failure of the X chromosomes to segregate properly during meiosis PZ. QA. Q: What (usually) determines the sex of a human embryo? QB. A: presence v. absence of the Y chromosome QC. QD. ____________________________: paring and synapse/ recombination; rest of the Y (male-specific region) has no counterparts on the X QE. QF. Q: What acts as a “switch” for expression of other genes? QG. A: ___________: (where other 74 genes are) contains genes, incl. sex-determining region Y (SRY)* present in all (male) mammals - SRY codes for the testis-determining factor (TDF) protein, a ‘switch’ for expression of other genes - SRY has a large effect by either facilitating or preventing transcription or translation of other genes QH. QI.Dosage Compensation • females have the potential to produce 2x as much protein encoded by X-linked genes than males • _______________________ mechanisms balance out the expression of X- linked genes - _____________ : inactivated X in diploid cells of females, visible during mitotic interphase QJ. QK. • Mechanism in silencing the other X chromosome is usually by _______________ - that X chromosome is never going to express any of its genes - this way, each gender has an equal amount of genes - Early on both chromosomes contribute to gene expression until one is silenced QL. QM. Lyon Hypothesis • one X chromosome per ___________ cell is randomly ___________ during embryonic development • ________________ have the same X inactivated as their parent cell, but not all parent cells are the same QN. QO. Temperature Dependent sex Determination QP. • Not like that across tree of life – sex may not even be genetically determined QQ. - Most common in reptiles that incupation temp sets the path whehter indivudual will be male or female QR.
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'