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Lectures posted online after lecture. Textbook sections and/or pages posted a few days prior to each lecture.
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In many instances there is a unique pattern of inheritance. Traits disappear and reappear in new ratios. CB 14.3
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GenotypePhenotype CB 14.6
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Human blood types AA or AO AB BB or BO OO CB tbl 14.2
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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?
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Mom=B blood BB or BO Dad=A blood AA or AO Gametes all B / 50% B and 50% O all A / 50% A and 50% O possible genotypes Frank can be BO = B blood …no worries
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We can also predict the future Fig 2.12
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Mom = ABDad = AB Inheritance of blood types
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Mom = ABDad = AB Gametes:A or B Inheritance of blood types
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Mom = ABDad = AB Gametes: A or B AA ABBB AB Chance of each phenotype for each offspring 25% AA 50% AB 25% BB Mom Dad Inheritance of blood types
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CB 14.7 Testcross: determining dominant/ recessive and zygosity
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CB 5.21 Sickle-cell anemia is caused by a point mutation
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Sickled and normal red blood cells
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Mom = HSDad = HS H or S HH HSSS HS possible offspring 75% Normal 25% Sickle-cell Mom Dad S=sickle-cell H=normal Sickle-Cell Anemia: A dominant or recessive allele?
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Coincidence of malaria and sickle-cell anemia CB 23.13
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Mom = HSDad = HS H or S HH HSSS HS possible offspring Oxygen transport: 75% Normal 25% Sickle-cell Malaria resistance: 75% resistant 25% susceptible Mom Dad Sickle-Cell Anemia: A dominant or recessive allele? S=sickle-cell H=normal
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Variation in Peas
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Phenotype Genotype CB 14.8
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The inheritance of genes on different chromo- somes is independent.
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Y y rR Gene for seed color Gene for seed shape Approximate position of seed color and shape genes in peas Chrom. 1/7Chrom. 7/7
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CB 15.2 The inheritance of genes on different chromosomes is independent: independent assortment
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CB 15.2 meiosis I meiosis II
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CB 15.2 The inheritance of genes on different chromosomes is independent: independent assortment
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CB 14.8
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Inheritance can be predicted by probability CB 14.9
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Probability of a 4= 1/6 Probability of two 4’s in a row= 1/6x1/6=1/36
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Probability of 3 or 4 = 1/6+1/6= 1/3
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“and” multiply “or” add
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Huntington’s Disease D=disease d=normal Neurological disease, symptoms begin around 40 years old.
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Mom = ddDad = Dd d or d D or d Dd dd possible offspring 50% Huntington’s 50% Normal Mom Dad Huntington’s Disease D=disease d=normal
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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)
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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
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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
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CB 14.11 Many traits are coded for by more than one gene.
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Eye color: One trait controlled by multiple genes
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Lectures posted online after lecture. Textbook sections and/or pages posted a few days prior to each lecture.
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