What happens when you examine how two traits are passed from one generation the next.
- P generation- true breeding
- yellow, round YYRR
- true breeding
- green, wrinkled yyrr
- F1 generation
- yellow, round. YyRr
- yellow is dominant to green, round is dominant to wrinkled.
- Y=yellwo
- R=round
- y=green
- r=wrinkled
- F2 generation
- YyRr x YyRr
- dihybrid cross
YR Yr yR yr YR YRYR YRYr YRyR YRyr Yr YrYR YrYr YryR Yryr yR yRYR yRYr yRyR yRyr yr yrYR yrYr yryR yryr - To do a dihybrid cross, d the monohybrid cross for each trait
- seed color Yy x Yy
Y y 1/4YY Y YY Yy 1/2Yy y Yy yy 1/4yy - Genotype ratio- ratio of genotypes in the offspring
- phenotype ratio - ratio of phenotypes in the offspring.
- seed shape Rr x Rr
R r R RR Rr r Rr rr - How do we put the mono-hybrid crosses together>
- Product Law 12:00 - probability of two events occurring simultaneously is equal to the product of the individual probability
- multiply the individual probabilities
- 3/4 are yellow multiplied by
- 3/4 round = 9/16 yellow and round
- 1/4 wrinkled = 3/16 yellow, wrink
- 1/4 green multiplied by
- 3/4 round = 3/16 green and round
- 1/4 wrinkled = 1/16 green wrinkled
- 9 yellow, round; 3 yellow, wrinkled; 3 green round; 1 green wrinkled
- Mendels 4th postulate (18:00)
- during gamete formation segregating pairs of unit factors (genes) assort independently of each other. Law of independent assortment.
- Trihybrid cross: watching 3 traits be passed from generation to generation (call the genes ABC)
- P generation (21:10)
- AABBCC x aabbcc
- F1 generation
- anytime you cross two homozygous individuals (one dominant, one recessive) you get a heterozygous for all three traits; AaBbCc
- F2 generation
- cross two heterozygotes; AaBbCc x AaBbCc
- do the three monohybrid crosses for Aa x Aa; Bb x Bb; Cc x Cc
A a 1/4AA 3/4A_ A AA Aa 1/2Aa a aA aa 1/4aa 1/4aa - genotype ratio
- 1/4 AA multiplied by
- 1/4 BB = 1/16 AABB multiplied by
- 1/4CC = 1/64 AABBCC (homozygous dominant for all three traits)
- 1/2Cc = 1/32 AABBCc
- 1/4cc = 1/64 AABBcc
- 1/2Bb (heterozygous) = 1/8 AABb multipied by
- 1/4 CC = 1/32 AABbCC
- 1/2Cc = 1/16 AABbCc
- 1/4cc =1/32AABbcc
- 1/4 bb (homozygous recessive) = 1/16 AAbb multiplied by
- 1/4 CC = 1/64 AAbbCC
- 1/2 Cc = 1/32 AAbbCc
- 1/4 cc = 1/64 AAbbcc
- 32 min explains what he looks for on tests
- Phenotype ratio is easier than genotype ratio - two phenotypes and three genotypes for each trait
- Probablilty laws influence genetic outcome
- Product law: The likelihood of two or more events occurring simultaneously is equal to the product of the individual probabilities.
- ex. 1/4 chance blue eyes and 1/8 chance of blonde hair gives you a 1/32 chance of having blue and blonde hair
- Binomial Theroum (38:55)
- can be applied to any situation where there are only 2 outcomes (ex. You are either male or female, right handed or left handed)
- you can multiply out probabilities over a number of events.
- You can use this for family composition. If you have "x" number of children what is the likelihood you will have "x" number of boys and "x" number of girls
- Bob and Judy want all boys what is the likelihood of this happening?
- (a+b)^n . . . . (a+b)^3
- general --> a^3 + a^2b + ab^2 + b^3
- for this problem --> 1a^3 + 3a^2b + 3ab^2 + 1b^3
- 1a^3 - 3 girls; 3a^2b - 2girls, 1 boy; 3ab^2 -2 boys, 1girl; 1b^3 - 3 boys
- (1/2)^3 = There is a 1 in 8 chance they will have all boys
- If they waned 2 boys and a girl
- 3(1/2)(1/2)^2
- 3 in 8 chance of 2 boys and 1 girl
- Jen and ben want a boy. They want 5 kids. If she has 5 kids what is the likelihood shell have at least one boy?
- 1a^5 5a^4b 10a^3b^2 10a^2b^3 5ab^4 1b^5
- A quick way to do this is to calculate the probability of having 5 girls (1a^5).
- (1/2)^5. . . 1 in 32 chance you'll have 5 gilrs. Therefore 1-(1/32)=31/32 chance of having at least one Boy
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