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Lectures posted online after lecture. Textbook sections and/or pages posted a few days prior to each lecture. CB 14.3 In many instances there is a unique pattern of inheritance. Traits disappear and reappear in new ratios. CB 14.6 Genotype Phenotype Human blood types AA or AO BB or BO AB OO CB tbl 14.2 Looking at the past: If Frank has B blood type, his Dad has A blood type, And his Mom has B blood type… Should Frank be worried? possible genotypes Mom=B blood Dad=A blood BB or BO AA or AO Gametes all B / 50% B and all A / 50% A and 50% O 50% O Frank can be BO = B blood …no worries We can also predict the future Fig 2.12 Inheritance of blood types Mom = AB Dad = AB Inheritance of blood types Mom = AB Gametes: A or B Dad = AB A or B Inheritance of blood types Mom = AB Gametes: A or B A or B Dad A or B A AA Mom or B AB Dad = AB AB BB Chance of each phenotype for each offspring 25% AA 50% AB 25% BB Testcross: determining dominant/ recessive and zygosity CB 14.7 Sickle-cell anemia is caused by a point mutation CB 5.21 Sickled and normal red blood cells Sickle-Cell Anemia: A dominant or recessive allele? Mom = HS Dad H or S H HH Mom or S HS HS SS S=sickle-cell H=normal Dad = HS possible offspring 75% Normal 25% Sickle-cell Coincidence of malaria and sickle-cell anemia CB 23.13 Sickle-Cell Anemia: A dominant or recessive allele? Mom = HS Dad H or S H HH Mom or S HS HS SS S=sickle-cell H=normal Dad = HS possible offspring Oxygen transport: 75% Normal 25% Sickle-cell Malaria resistance: 75% resistant 25% susceptible Variation in Peas Phenotype Genotype CB 14.8 The inheritance of genes on different chromosomes is independent. CB 14.8 Approximate position of seed color and shape genes in peas Y y Gene for seed color r Chrom. 1/7 R Chrom. 7/7 Gene for seed shape The inheritance of genes on different chromosomes is independent: independent assortment CB 15.2 CB 15.2 meiosis I meiosis II The inheritance of genes on different chromosomes is independent: independent assortment CB 15.2 CB 14.8 CB 14.9 Inheritance can be predicted by probability Probability of a 4= 1/6 Probability of two 4’s in a row= 1/6x1/6=1/36 Probability of 3 or 4 = 1/6+1/6= 1/3 “and” multiply “or” add Huntington’s Disease D=disease d=normal Neurological disease, symptoms begin around 40 years old. Huntington’s Disease Mom = dd Dad = Dd Dad D or d d Dd Mom or d Dd D=disease d=normal dd dd possible offspring 50% Huntington’s 50% Normal Two different people: One with Huntington’s disease = Dd Hh One without Huntington’s disease = dd Hh mate. What is the probability that their offspring will have Huntington’s disease and sickle cell anemia? (Dd hh) Two people: One with Huntington’s disease = Dd Hh One without Huntington’s disease = dd Hh mate. What is the probability that their offspring will have Huntington’s disease and sickle cell anemia? Probability of each outcome: Probability of Dd (Ddxdd) = .5 Probability of hh (HhxHh) = .25 Dd hh Two people: One with Huntington’s disease = Dd Hh One without Huntington’s disease = dd Hh mate. What is the probability that their offspring will have Huntington’s disease and sickle cell anemia? Dd hh Probability of each outcome: Probability of Dd (Ddxdd) = .5 Probability of hh (HhxHh) = .25 Multiply both probabilities .25 X.5 = 12.5% chance Dd hh offspring Many traits are coded for by more than one gene. CB 14.11 Eye color: One trait controlled by multiple genes Lectures posted online after lecture. Textbook sections and/or pages posted a few days prior to each lecture.