MBG 2040 Midterm Review Notes.docx

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Molecular Biology and Genetics
MBG 2040
T.Ryan Gregory

MBG 2040 Midterm Review Notes Topic 1 4 basic patterns of inheritance for single gene traits • Autosomal recessive • Autosomal dominant • Sex linked recessive • Sex linked dominant Humans: 2N=46 xsomes (23 pairs of chromatids makeup 46 chromosomes) o 22 pairs of autosomes o 1 pair of sex chromosomes (xx or xy) xx chromosome has 1000 + genes xy chromosome has 300~ genes Autosomal Traits Recessive ex. Classic albinism, sickle cell anaemia, cystic fibrosis AA normal Aa carrier Aa affected Dominant ex. Huntington disease, polydactyly, neurofibromatosis BB affected Bb affected Bb normal • Trait controlled by Rare Dominant Allele, the carrier is hetero because the chances of producing offspring with trait BB is ¼ since (Bb x Bb = ¼ BB) Ex. Widows peak, eye colour, lips thick Structural Gene |Regulatory Region| |Protein Encoding Region| DNA Regulatory Region: determines when and how often transcription occurs Protien Encoding Region: specifies the amino acid sequence of the protein RNA polymerase: binds to promoter site Transcription factors bind to DNA Recessive Mutations: often involve loss of gene function Null or Amorphic Alleles |Regulatory Region| |Protein Encoding Region| no transcription protein has no activity DNA strand Results: i) protein present but not functional OR ii) no protein produced Hypomorphic Allele |Regulatory Region| |Protein Encoding Region| Reduced transcriptionprotein with reduced activity Results: i) reduced amounts of normal protein OR ii) altered protein with reduced acitivity Himalayan and Siamese Cats are homozygous for a hypomorphic allele of the tyrosinase gene Allele Where On DNA Strand Phenotype A normal wild type a 1 protein encoding region severe mutant phenotype a 2 in regulatory region severe mutant phenotype a 3 protein encoding region produces partially functional polypeptide mild mutant phenotype a 4 protein encoding region produces reduced amounts of polypeptide mild mutant phenotype Autosomal Recessive Pedigree determination: ­ Heterozygosity value reduced by ½ with each AA mating ­ Marrying in always homo dominant unless stated differently Offspring between Aa x AA > ½ Aa ½ AA Offspring between Aa x Aa > 2/3 Aa 1/3 AA Determining crosses… P1 ½ AA ½ AA mated to Aa what is offspring? 1. ½ Aa Aa 2. ½ AA Option 1: ½ Aa x Aa x ?(Aa) = () what we are looking for….must look at punnet square possibilities which is 1/3 AA 2/3 Aa therefore: ½ Aa x Aa x 2/3 (Aa) = 1/3 possibility of being Aa (which is an a allele carrier) repeat process for option 2 and sum the totals…. = 7/12 Aa Labeling a autosomal recessive pedigree Step 1: assign all affected individuals (aa) genotypes Step 2:label all affected individuals parents (Aa) genotypes Step 3: label anyone marrying in (AA) genotypes **Remember** Aa x AA offspring are ½ Aa ½ AA Aa x Aa offspring are 2/3 Aa 1/3 AA Autosomal Dominant Traits Dominant Mutations: often involve gain/change of gene function Dominant Hypermoprhic Alleles: |Regulatory Region| |Protein Encoding Region| Incr. transcription protein with incr. activity DNA Results: i) overproduction of normal protein ii) production of protein with incr. activity Neomorphic Alleles: |Regulatory Region| |Protein Encoding Region| DNA Protein with new function Results: i) production of an altered protein with new function ex. Ehlers Danlos Syndrome is a collagen disorder : hetero for dominant neomorphic allele Autosomal Dominant Pedigrees: - Rare and dominant traits are usually heterozygous individuals (Bb) - Unaffected individuals are homozygous recessive (bb) ex. Bb x Bb have offspring ¼ BB ½ Bb ¼ bb>>>>> 2/3 Bb 1/3 BB affected offspring Bb x Bb x ¾ (B-) = ¾ to have affected offspring since ¾ automatically affected Q) PKU autosomal recessive disease.. a normal man marries a normal woman from outside the family and consider having a child, whats probability child will have PKU? What if the frequency of PKU carriers is 1 in 50…? 1/50 Pp (chance) x Pp (man or woman) x ¼ (pp) (punnet square) = 1/200 o The chance their first child will have PKU is 1/200. 1/50 Pp x 1/50 Pp x ¼ pp = 1/10000 o The chance 2 unrelated normal people from this population have a PKU child is 1/10000. Q) What are chances these 2 normal people with have a normal child? i) criss cross option 1/50 Aa x 1/50 Aa = 9,999/10,000 49/50 AA x 49/50 AA ii) subtract probabaility of being affected from 10 000 (1-10 000 = 9999/10 000) Hardy Weinberg Equation i) p + 2pq + q = 1 ii) Frequency of CF is 1/2500 iii) p= frequency of recessive c allele iv) q= C allele frequency p = 1/2500 p= 0.02 (c allele)and since p + q =1 we know 1- 0.02 = q (0.98) (C allele) 2pq=0.02(0.98)= 0.0392 Topic 2 Chromosomes, Sex Determination and Sex Linkage Chromosomes made up of o chromatin (complex DNA and Proteins) o dynamic pieces alternating between tight and loose compaction o occupy nucleus regions called “chromosome territiories” o can be visualized when tightly compacted Karyotype o number and appearance of xsomes from eukaryotic somatic cell o xsomes must be compacted and cells used for karyotyping must be mitotic o catch xsome in mitosis when compacted not meiosis since not visual (testes…) o xsomes have been arranged in pairs o cells that have been arrested in mitosis and each chromosome consists of two identical sister xtids Giesma: staining bands on xsomes, dark bands mean heterochromatic regions Euchromatin o less condensed o gene rich o at xsome arms o recombination at meiosis Constitutuve Heterochromatin: like y chromosome in males o highly condensed o gene poor o at centromeres and telomeres o no meiotic recombination - Y xsome encodes testis determining factor at SRY gene Birds: females ZW males ZZ Turtles: sex dependent on heat in incubation Ants, Bees, Wasps females diploid (fertilized egg); males haploid (unfertilized egg) Drosophila 2N=8 males determining genes autosomes, female x chromosome • 6 autosomes + XX fertile female • 6 autosomes + XY fertile male • 6 autosomes + X sterile male • 6 autosomes + XXY fertile female Humans: • XO 45 xsomes not 46 – female with turner syndrome • XXY 47 xsomes not 46- male with klinefelter syndrome Drosophila: eye color maps to x chromosome ­ Red eye males and females are common (R) ­ White eyed males were rare (W) ­ R female x W male F1 red eyes F2 all females R ½ males R ½ males W This leads to X linked traits ­ Red green colorblindness ­ Males are hemizygous, one allele is Y chromosome therefore only X can be affected o males give their x allele always to their daughters.. if she shows traits of not have fathers phenotype/genotypes then its not x linked o ex. XBY x XbXb daughter XbXb not affected… therefore autosomal dominant ­ Haemophilia blood clotting disorder ­ Dystrophin cytoskeleton in proteins in membrane ­ Barr Body darkly staining structure inside nucleus(females have, males don’t) o XO females have no Barr Bodies o XXY males have 1 Barr Body o XXXY females have 2 Barr Bodies Barr Body is an inactivated X chromosome, xsomes of this aren’t being transcribed ­ Consists of highly compacted chromatin (heterochromatin) ­ Maternal X active, paternal barr body: descendants of this cell will inactivate paternal X xsome ­ Paternal X active, maternal barr body: descendants of this cell will inactivate maternal x xsome Mammal females are mosaics: ­ mixture of the two cell types, (maternal x active, paternal x active) ­ ex calico, tortoise shell cats Males: XOb Yblack males XOo Y orange yellow males Females XOb XOb black female XOo XOo orange yello female XOb XOo tortoiseshell calico female Some cells have X xsome with black allele inactivated= orange spots Some cells have X xsome with orange allele inactivated= black spots Topic 3 Linkage 3 possible scenarios for a dihybrid test cross i) genes A and B Independently assort a. 4 classes of offspring with equal frequencies ii) genes A and B tightly linked (no crossing over) a. 2 classes of offspring with equal frequencies (cis trans) iii) genes A and B linked, crossing over occurs between A and B a. 4 classes of offspring with 2 large and 2 small frequencies i. 2 parental ii. 2 recombinants Scenario 1. Genes A and B independently assort Dihybrid AaBb x aabb Gametes from AaBb : AB Ab aB ab Therefore gametes combine of the dihybrid aabb: ab parent and test cross parent to give…. AB Ab aB ab ab ab ab ab 4 classes of offspring with equal frequencies Scenario 2: Genes A and B located closely on same chromosome..impossible for crossing over A A a a B B b b Dihybrid Alleles (AaBb) in cis configuration AAlleles in trans configuration aB ab Ab Now do test cross with cis or trans... this ex. in cxs AB ab Ab ab Offspring: AB ab Therefore two classes of offspring with equal frequencies ab ab Scenario 3. Genes A and B are tighly linked and crossing over occurs between A and B A A a a Ab and aB on outside chromosome not b b B B involved in crossing over A a A a Ab aB parentals not in x over B b B B ab AB recombinants produced by x over Dihybrid Test Cross aB ab Ab x ab Gametes: aB parental ab only aB/ab parental150 Ab parental Ab/ab parental 150 ab recombinants ab/ab recomb 50 AB recombinants AB/ab recomb 50 • Number off recombinants/ total offspring x100 = map units ex. 100/400 x 100= 25 map units b/w A and B genes • Independent assortment can imply that two genes are on different chromosomes OR o 50 or more map units apart on same chromosome 4 gametes after meiosis crossing over: they tell you the recomb is 10.8% • That means 10.8 / 2 = each recomb gamete • 100-10.8= parental frequencies • 89.2/ 2 = each parental gamete frequency Parental Parental Recombinant Recombinant 44.6% 44.6% 5.4% 5.4% Nail Patella Syndrome linked to ABO blood phenotype Type A blood Type B blood Type AB bloodType O blood IA IA IB IB IA IB ii IA i IB i Haemophila and red green color blindness are linked ­ first assign genotypes to males, since have y xsome ­ males x xsome is their phenotype so that’s assumed ­ h and cb are haemophila and color blindness (homozy recessive) ­ fathers pass down their x chromosome to daughters always ­ look at what offspring are, that determines other x xsome for mom Probability of producing a child who is CH b / Yxsome ­ prob of y xsome from dad X prob of CBh from mom= ­ from her recomb gene theres parental alleles 0.10 x 0.45=0.045 ­ from her parental gene theres parental and recomb allleles o 45% and 5% so 50% total…0.90x 0.50=0.45 Overall back to equation now… 0.5 from dad X CHb (0.045 x 0.45) from mom = 0.248 ­ the male offspring has a 0.248 chance of receiving CHb from parents Topic 4 Extensions of Mendelian Analysis Simple/Complete Dominance ­ BB & Bb dominant over bb causing phenotypic ratio to be different 9:7 etc Incomplete Dominance/Semi Dominance/Partial Dominance (all same) ­ The phenotype can be distinguished from the phenotype of both homozygotes ­ Hybrids don’t breed true> breed hetero (true is homozygous only) ­ Ex C+C+ sorrel, Ccr C+ palomino, Ccr Ccr cremello Multiple Alleles ­ Some genes have multiple alleles o Some c
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