MBG 2040 Midterm Review Notes.docx

15 Pages
223 Views
Unlock Document

Department
Molecular Biology and Genetics
Course
MBG 2040
Professor
T.Ryan Gregory
Semester
Fall

Description
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
More Less

Related notes for MBG 2040

Log In


OR

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Sign up

Join to view


OR

By registering, I agree to the Terms and Privacy Policies
Already have an account?
Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.

Add your courses

Get notes from the top students in your class.


Submit