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Exam #3 W 12/5 at 7-8:30pm in ETC 2. 108 for the 9am class and ECJ 1
Exam #3 W 12/5 at 7-8:30pm in ETC for the 9am class and ECJ for the noon class Review T 12/4 at 5pm in WRW 102 Homework #4 due 12/3 (if needed) Bonus #2 is posted
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Grades: A = 88.5+ B = 77.5 C = 65.5 Exam avg x homework + bonus = grade
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Tracking two separate genes, for two separate traits, each with two alleles.
Ratio of 9:3:3:1 Fig 3.4
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Box 2.2 Crossing-over Meiosis: In humans, crossing-over and independent assortment lead to over 1 trillion possible unique gametes. (1,000,000,000,000) Meiosis I (Ind. Assort.) Meiosis II 4 Haploid cells, each unique
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Crossing over produces new allelic combinations
Fig 4.3
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Homologous pair of chromosomes
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Longer regions have more crossovers and thus higher recombinant frequencies
Fig 4.10
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Some crosses do not give the expected results
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=25% 42% 41% 9% 8%
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These two genes are on the same chromosome
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By comparing recombination frequencies, a linkage map can be constructed
= 17 m.u.
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Another test
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42 recombinants out of 381 offspring =
42/381 11% recombination
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The probability of crossing over can be used to determine the spatial relationship of different genes Fig 4.9
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Double recombinants arise from two crossovers
Fig 4.11
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Double recombinants can show gene order
Fig 4.12
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What is the relationship between these 3 genes
What is the relationship between these 3 genes? What order and how far apart? similar to pg 141
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What is the relationship between these 3 genes
What is the relationship between these 3 genes? What order and how far apart? similar to pg 141
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Double crossover similar to pg 141
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Which order produces the double crossover?
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Which order produces the double crossover?
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We have the order. What is the distance?
similar to pg 141
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Recombinants between st and ss: (50+52+5+3)/755 =14.6%
similar to pg 141
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Recombinants between ss and e: (43+41+5+3)/755 =12.2%
similar to pg 141
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Put it all together… 26.8 m.u. st ss e 14.6 m.u. 12.2 m.u.
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Linkage map of Drosophila chromosome 2
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Recombination is not completely random.
physical distance linkage map Yeast chromosome 3 Fig 4.13 and 20
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It is rarely this simple.
Genotype Phenotype Genes code for proteins (or RNA). These gene products give rise to traits… It is rarely this simple. Figs
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The relationship between genes and traits is often complex
Complexities include: Complex relationships between alleles Multiple genes controlling one trait One gene controlling multiple traits Environmental effects
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The colors of peppers are determined by the interaction of several genes
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Complexity of inheritance leads to genetic diversity.
Fig
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Eye color: One trait controlled by multiple genes
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Seven alleles and their interactions in leaf patterning of clover
Fig 6.7
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The relationship between genes and traits is often complex
Complexities include: Complex relationships between alleles Multiple genes controlling one trait One gene controlling multiple traits Environmental effects
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1 gene controlling many traits
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Mom = HS Dad = HS Dad H or S possible offspring 75% Normal
S=sickle-cell H=normal Sickle-Cell Anemia Mom = HS Dad = HS Dad H or S possible offspring 75% Normal 25% Sickle-cell HH HS H or S Mom HS SS
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Coincidence of malaria and sickle-cell anemia
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Mom = HS Dad = HS possible offspring Oxygen transport: 75% Normal
S=sickle-cell H=normal Sickle-Cell Anemia Mom = HS Dad = HS possible offspring Oxygen transport: 75% Normal 25% Sickle-cell Malaria resistance: 75% resistant 25% susceptible Dad H or S HH HS H or S Mom HS SS
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The relationship between genes and traits is often complex
Complexities include: Complex relationships between alleles Multiple genes controlling one trait One gene controlling multiple traits Environmental effects
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Next we will look at what DNA can tell us about the origins of Homo sapiens.
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Exam #3 W 12/5 at 7-8:30pm in ETC 2. 108 for the 9am class and ECJ 1
Exam #3 W 12/5 at 7-8:30pm in ETC for the 9am class and ECJ for the noon class Review T 12/4 at 5pm in WRW 102 Homework #4 due 12/3 (if needed) Bonus #2 is posted
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