Section 14. Pedigree Analysis and MolecularMarkersBasic Pedigree AnalysisIn humans, one mating (one pair of parents) rarely has enough children togive reliable ratios. In that case one can still do pedigree analysis. Modernpedigree analysis is much more sophisticated than anything we can do; wewill use pedigrees mainly as good practice in basic genetic analysis.Mendel could control his crosses, mate any pea plant with any other. Wehave 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 constructpedigree.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-downintroduction to genetics, some students in previous classes have used Schaum’s Outline ofGenetics. 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 examto us today or next week and we will give you 3 points. This is because an error in thelecture Section 11 on Sex and Meiosis would lead you to give that answer. The correctedversion of Section 11 is on the web. The corrected sequence of events in male and femaleanimal 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 oneegg.This will increase the mean score somewhat so I won’t post the statistics until we havecorrected 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 arehomozygous recessive. Fill in genotypes in steps:(1) All affected are homozygous recessive.Try simplest hypothesis first: 1 gene, 2 alleles, complete dominance, affected arehomozygous 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 arehomozygous 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 arehomozygous 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 arehomozygous 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. withoutcontradicting our hypothesis.The genotype of I2 can be AA or Aa. Can we deduce that it is AA because it had no aaoffspring?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 sizeis too small.However, one can calculate the probabilities that I 2 is AA or Aa, taking into account theinformation from the offspring. Important in genetic counseling, where I 2 may want toknow the probability that her next offspring will be affected. Uses a method ofconditional probability called Bayes' theorem. The conditional probability that I 2 ishomozygous 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 todominant allele D, normals dd:Problem: the identical twins in generation II, and III 2 and 3, must have dominantallele, must have got from one or other parent, so one of their parents would haveto be albino. Internal contradiction.A major problem in studying and treating human hereditary diseases is our inability to identifyheterozygous carriers of recessive genetic defects. Need to do so to counsel them about havingchildren. (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 willdevelop 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 notappear before individuals reach reproductive maturity. Show incomplete penetrance = failureto 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 0.350 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 populationhave a particular restriction site; others lack it.Can be due to difference in single base pair (single nucleotide polymorphism = SNP), or toinsertions or deletions.Now to see whether individual is AA, Aa, 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 fragmentsAA 0.9Aa 0.9, 0.5, 0.4aa 0.5, 0.4.Mutations that are due to a transposable element can be detected by PCR of the region andchecking its size on a gel.Mutations that are due to a transposable element can be detected by PCR of the region andchecking its size on a gel. E.g. wrinkled allele in peas, white