PowerPoint PresentationSection 14. Pedigree Analysis and Molecular MarkersBasic Pedigree AnalysisSlide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17CODIS - combined DNA index systemThis presentation was originally prepared by C. William Birky, Jr.Department of Ecology and Evolutionary BiologyThe University of ArizonaIt may be used with or without modification for educational purposes but not commercially or for profit.The author does not guarantee accuracy and will not update the lectures, which were written when the course was given during the Spring 2007 semester.Section 14. Pedigree Analysis and Molecular MarkersBasic Pedigree AnalysisIn 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 "natural experiments":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 20,000 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-> primary spermatocytes –MI-> secondary spermatocytes -MII-> spermatids -> differentiate-> spermoogonia –mitosis-> primary oocyte –MI-> secondary oocyte –MII-> ovum (egg) + polar body + polar bodyNOTE: 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 "marriage line".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.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.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.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.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.We have filled in the pedigree without finding any internal contradictions, i.e. without contradicting our hypothesis.The genotype of I2 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 Aa out of 4 offspring is (1/2)4 = 1/16 > 1/20 or 0.05, 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 that I 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 Bayes' theorem. The conditional probability that I 2 is homozygous is 0.6124; this is the probability that would be used in genetic counseling.Return to pedigree with albinism. Try with other hypothesis, with albinism due to dominant allele D, normals dd:Problem: the identical twins in generation II, and III 2 and 3, must have dominant allele, must have got from one or other parent, so one of their parents would have to be albino. Internal contradiction.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.e.g. Huntington's disease (Huntington's chorea):age penetrance = proportion of individuals who are known to carry the genotype for the disease and actually show symptoms by this age30 0.140 0.3 50 0.660 0.8570 0.95Molecular markers can solve these problems. Many ways of detecting polymorphic differences.Molecular MarkersRFLP = restriction fragment length polymorphism: some individuals in a population have a particular