Class Note for ECOL 320 at UA
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Date Created: 02/06/15
Section 14 Pedigree Analysis and Molecular Markers Ill TK Halimnon Basic Pedigree Analysis In humans one mating one pair of parents rarely has enough children to give reliable ratios In that case one can still do pedigree analysis Modern pedigree analysis is much more sophisticated than anything we can do we will use pedigrees mainly as good practice in basic genetic analysis Mendel could control his crosses mate any pea plant with any other We have to take advantage of quotnatural experimentsquot 1 Find individual with unusual trait whose inheritance is to be studied propositus 2 Examine as many relatives as possible for presence of trait and construct pedigree 3 Analyze to determine mode of inheritance Very important in medical genetics and genetic counseling For students who want an extra source of practice problems or another trimmed down introduction to genetics some students in previous classes have used Schaum s Outline of Genetics It is available from the bookstore on order delivery time one or a few days On the exam if you answered 20000 to the second part of question 15 return your exam to us today or next week and we will give you 3 points This is because an error in the lecture Section 11 on Sex and Meiosis would lead you to give that answer The corrected version of Section 11 is on the web The corrected sequence of events in male and female animal gametogenesis is spermatogonia mitosis gt primary spermatocytes Ml gt secondary spermatocytes Mll gt spermatids gt differentiate gt sperm oogonia mitosis gt primary oocyte Ml gt secondary oocyte MlI gt ovum egg polar body polar body NOTE This does NOT affect the answer to the first part one oocyte gives rise to only one egg This will increase the mean score somewhat so I won t post the statistics until we have corrected the grades Generations labelled roman numerals I II Individuals labelled arabic numerals 1 2 Males square females round Shaded affected Mated individuals connected by quotmarriage linequot ll 39 a E 3inuwiduais J identical twins III 131 II D E II I I I II E r fraternal twins Try simplest hypothesis rst 1 gene 2 alleles complete dominance affected are homozygous recessive Fill in genotypes in steps 1 All affected are homozygous recessive may lt gt Eilezmuwmuels aaaa menueel twms lt as ltaalt 1raterna twms Try simplest hypothesis rst 1 gene 2 alleles complete dominance attetteel ate homozygous recessive Fill in genotypes in steps 1 All atteeteel ate homozygous recessive 2 All unaffected have at least one dominant allele 63mdwldua s A aaaa A H El El A AAA marinas twms A as as A Yraterna twms Try simplest hypothesis first 1 gene 2 alleles complete dominance affected are homozygous recessive Fill in genotypes in steps 1 All affected are homozygous recessive 2 All unaffected have at least one dominant allele 3 All homozygous recessive must get one recessive allele from each parent 2 l O 13 3 A 4 m 339 shall Aai Ell g E A I A i A i Iiita 0 Q 31mjiu1duais ma taa aai ta i 3 identical twins Hi i P ahl IIaaIl iaa xl 3 1 fraternal twina Try simplest hypothesis first 1 gene 2 alleles complete dominance affected are homozygous recessive Fill in genotypes in steps 1 All affected are homozygous recessive 2 All unaffected have at least one dominant allele 3 All homozygous recessive must get one recessive allele from each parent 4 All offspring of homozygous recessive must have at least one recessive allele 1 a 7 l J F 3 4 an A A A6 IZAEI39J II El III E f b l3 31ndmnua15 A3 ma AE1 Ma ail aal aali rm II xx K39s M 3 1Ient1cal twms Ill 6 U11 quot A II at Em A 339 RR fraternal tw1r13 Try simplest hypothesis first 1 gene 2 alleles complete dominance affected are homozygous recessive Fill in genotypes in steps 1 All affected are homozygous recessive 2 All unaffected have at least one dominant allele 3 All homozygous recessive must get one recessive allele from each parent 4 All offspring of homozygous recessive must have at least one recessive allele 5 For rest of genes use allele unknown 1 1 3 I39 H34 i39 r i J Ma ma r I III El i i i 3 i Bir39ldiuiduals A3 A311 A3 A51 may ail MathiasII lx 79 identical twins Ill a r 1 H LAD ea Laa lie1 7raternaltw1ns We have filled in the pedigree Without nding any internal contradictions ie Without contradicting our hypothesis The genotype of 12 can be AA or Aa Can we deduce that it is AA because it had no aa offspring NO if it is Aa the probability of getting 4 Au out of 4 offspring is 124 116 gt 120 or 005 too high to reject In general one cannot use ratios to determine genotypes in pedigrees because the sample size is too small However one can calculate the probabilities thatI 2 is AA or Aa taking into account the information from the offspring Important in genetic counseling where I 2 may want to know the probability that her next offspring will be affected Uses a method of conditional probability called Bayes39 theorem The conditional probability that I 2 is homozygous is 06124 this is the probability that would be used in genetic counseling 1 Cl l 1 J1 ij 1 4 Ea iii339 nail nail W I fa LJJ I El El 1 I 7 3 individuals Ana A3 AE1 mall ha aaiiaai IAII FR quotN identicai twins 2quot R IN I J Ex iiiTi an I166 iiiII 39 fraternal twins Return to pedigree with alhinism Try with other hypothesis with alhinism due to dominant allele D noimals dd Problem the identical twins in generation ll and lll 2 and 3 must have dominant allele must have got om one oi othei parent so one oftheir parentswould have to he alhino Internal contradiction 63mdwmua s lad DD marinas twms lt D low 1raterna twms Molecular Markers A major problem in studying and treating human hereditary diseases is our inability to identify heterozygous carriers of recessive genetic defects Need to do so to counsel them about having children Also problem in doing genetics with any diploid organism Many parents would like to use amniocentesis to find out if the embryo they are carrying will develop hereditary defects at a developmental stage when abortion is still an option Even some dominant defects cannot always be detected in time because symptoms may not appear before individuals reach reproductive maturity Show incomplete penetrance failure to be expressed in all individuals of the appropriate genotype eg Huntington39s disease Huntington39s chorea age penetrance proportion of individuals who are known to carry the genotype for the disease and actually show symptoms by this age 30 01 40 03 50 06 60 085 70 095 Molecular markers can solve these problems Many ways of detecting polymorphic differences RFLP restriction fragment length polymorphism some individuals in a population have a particular restriction site others lack it Can be due to difference in single base pair single nucleotide polymorphism SNP or to insertions or deletions region ampli ed w h a 313th FOR pllITIBIE I 0911 I 11111219 1 1 A 51351213 A13TEBEIZIZTATIZ quotAEBEET mutation 51 13MB quotAETIZEEViIECTATC quotAEBEET 1411 Hm I 0511 1 Now to see Whether individual is AA Au or as isolate a small sample of DNA amplify this region restrict the amplified DNA with HaeIII run on gel to separate fragments and estimate their sizes genotype fragments AA 09 Aa 09 05 04 an 05 04 Mutations that are due to a transposable element can be detected by PCR of the region and checking its size on a gel Mutations that are due to a transposable element can be detected by PCR of the region and checking its size on a gel Eg wrinkled allele in peas white in Drosophila In these cases the molecular marker is the size difference between the alleles that we are trying to detect More common case is when molecular marker is closely linked to the difference between our subject alleles Especially useful are DNA regions of tandem repeats of 29 nucleotides known variously as Omicrosatellites simple sequence repeat polymorphism SSRP simple tandem repeat polymorphism STRP 0 short tandem repeats STR Hypothetical example GACGACGACGAC GAC4 GACGACGACGACGAC GAOS Copy number varies greatly and size differences alleles can be detected by PCR of the region and separating PCR products on a gel Look for SSRP closely linked to gene and such that mutant allele of gene is closely linked to one SSRP allele while the normal allele is linked to another allele A repeat of 1060 nucleotides is called a variable number of tandem repeats VNTR Changes in copy number occur so often that are many different variants in population emunpm mu hm mwmcm mam Numbquot m can a musmm mm D214 awn Emaquot Wm Mumv I Forensic DNA Analysis If look at several different regions where there are SSRPs or VNTRs no two individuals are identical Use for forensic DNA analysis DNA fingerprinting DNA profiling DNA typing The Federal Bureau of Investigation FBI uses a standard set of 13 specific STR regions for the Combined DNA Index System CODIS CODIS is a software program that operates local state and national databases of DNA profiles from convicted offenders unsolved crime scene evidence and missing persons The odds that two individuals will have the same 13loci DNA profile is about one in one b39ll39 1 ion New Use 13 primer pairs to amplify different microsatellite regions Machine separates fragments and gives profile of peaks corresponding to different alleles W will llVIHIIII gt Ily combined DNA index system Share DNA profiles among crime labs 1994 official US government approval and standards set 112005 had 124200 forensic rofild E with Chromosomal Positions 2770 matches made thus far Z quot
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