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MCDB 2150Exam # 1 Study Guide Lectures: 1 - 11Lecture 1 (1/12)Syllabus dayLecture 2 (1/14)Know chromosome structure:Somatic cells: exact duplicates of each other (except immune system)Mitosis: Process of somatic cell divisionLecture 3 (1/16)Gene – a unit of heredityAllele – alternate form of a single gene (dominant/recessive versions, etc.)Homozygous – two of the same allelesHeterozygous – different allelesPunnett Square:Representation of possible gametes from each parentSimple cross: Dihybrid cross (2 genes= R & F)Finding probability using product rule:For the probability of two (or more) independent events occurring togetherFormula:P of event 1 x P of event 2 x P of event 3 x …..Finding probability using the addition rule:Probability of mutually exclusive events (OR events)Formula:P of event 1 + P of event 2 + P of event 3 …Lecture 4 (1/21)Chi Square Analysis:Step 1: Collect dataStep 2: Form hypothesisStep 3: Calculate expected(probability of first phenotype)x(total) =(probability of second phenotype)x(total) =Step 4: Apply Chi Square formula:Do for each phenotype separately ^^Add together all answers for the X^2 valueStep 5: Find P value using X^2 value & chart- Degrees of freedom = # of phenotypes – 1o 2 phenotypes (degree of freedom = 1)o 3 phenotypes (degree of freedom = 2)- plug in X^2 value in correct degree of freedom & read corresponding P valueStep 6: accept/reject hypothesisIf P value is MORE than 0.05 = not significant difference = accept hypothesisIf P value is LESS than or equal to 0.05 = significant different = reject hypothesisLecture 5 (1/23)Autosomal chromosomes: Normal (non-sex determining) chromosomesSex determination:Chromosomal- Homogametic (xx = female) & heterogametic (xy = male)- Dosage (of x) (in some invertebrates)Non chromosomal- Environmental- Scattered genes (not on designated chromosome)- Temperature can determine sex (affect on hormone levels)- Some organisms can change sexPossible genotypes for X Chromosome (individual lives– not so for most autosomal chromosomes)x, xxx, yxx, (some more too)BECAUSE = BARR BODIESExtra X chromosomes will be inactivated Throughout a female organism different Xs could be inactivated showing different phenotypesLecture 6 (1/26)Pedigrees!!- Dominant traits will show affected individuals in every generation- Recessive traits can go many generations without affected individuals- Circle = female- Square = male- White = unaffected- Black = affected- White/black = carrier- X-linked recessive traits will typically only show affected malesMitochondrial inheritance- Mitochondria also have their own DNA- Deposited in zygote by the egg cell (from mother)Lecture 7 (1/28)Penetrance = # or % of individuals with the genotype that show the expected phenotype (QUANTITY)Expressivity = the degree or intensity of a phenotype (QUALITY)Lecture 8 (1/30)**Heterogeneous trait = a trait that can arise from a mutation in any number of different genes (aka genetic heterogeneity)Complementation = mutation is on the SAME gene** to find gene groups: make a list of which mutants complement each other**know that ratio^^^^Maternal effect:Embryo uses the mother’s gene products (in egg) for the first few hours of development The child will ALWAYS show the phenotype of the mother’s genotype (regardless of theirs)Mother’s phenotype is not affected by her genotype in this caseLecture 9 (2/2)Independent assortment: two genes on different chromosomesRecombination probability of two genes is 50%When the genes are:Far apart – more likely to recombineClose together – less likely to recombineLinked chromosomes = do NOT assort independently (no mixing)Unlinked chromosomes = assort independently (50% of the time)Calculate distance of genes on a linked chromosomes (2 point cross)(# of recombinants)/(total offspring) x 100 = centimorgans (cM)Step 1: Interpret data which offspring are non-crossovers=NCO (more common)which are single cross overs?= SCO (way less common)Step 2: Determine parental haplotypes from the NCOsStep 3: Numerator = all SCOsDenominator = total # of offspringMultiply the whole thing by 100 (the formula)Lecture 10 (2/4)3 Point cross tests(# of recombinants)/(total offspring) x 100 = centimorgans (cM)Step 1: Interpret data which offspring are non-crossovers=NCO (more common)which are single cross overs?= SCO (less common)Which are double cross overs? = DCO (even less common)Step 2: Determine parental haplotypes from the NCOs & DCOsLay out the two known parental gametes from NCOs & DCOs – compareThe 1 gene that switches or the 1 gene that stays the same = middle gene!Step 3: Find distance between genesIf its between the middle and an end gene,Numerator: plug in the SCOs associated with this crossPlug in all DCOsDenominator = total # of offspringMultiply the whole thing by 100 (the formula)If its between the two end genesNumerator: plug in ALL SCOs Plug in all DCOs x2 = 2(3+6)Denominator = total # of offspringMultiply the whole thing by 100 (the formula)Lecture 11 (2/6)Review for


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