Unformatted text preview: ances it is a heart]  (1/13x1/4)
[Chances it is not both] = Important Special Case of the Sum Rule: If A and B are mutual exclusive events, the probability one of
them occur, P(A or B) is P[A + B] = P(A) + P(B) if you roll a dice, what is the probability of getting a 5 or 6? (1/6) + (1/6) = 1/3 Probability • What is the probability of drawing an Ace or a King from this deck of cards? 1/13 (getting an Ace) + 1/13 (getting a King) = 2/13 IV. Principle of Independent Assortment in
a Dihybrid Cross Monohybrid = one gene= heterozygote = Aa Dihybrid = two genes = heterozygous for both genes = AaBb Dihybrid Cross: T genes
wo
Why is it a pure breeding line? Homozygous X P
RRyy round green rrYY wrinkled yellow F1
RrYy round yellow F2 315 round yellow
108 round green
101 wrinkled yellow
32 wrinkled green The thing for his success is the fact he
looked for the ratio which is 9:3:3:1. Dihybrid Cross → 9:3:3:1 Phenotypic Ratio
RrYy X RrYy 315 round yellow
9
108 round green
3
101 wrinkled yellow
3
32 wrinkled green
1
556 seeds
16 RYRyy
rrYrryy homozygous recessive • Mendel added up the numbers of individuals in certain F2 phenotypic
classes to determine if the monohybrid 3:1 F2 ratios were still present. round : wrinkled
315 + 108 : 101 + 32 yellow : green
315 + 101 : 108 + 32 423 : 133 416 : 140 3:1 3:1 Explanation of the 9:3:3:1 • The 9:3:3:1 ratio of a dihybrid cross can be predicted because we
can consider each trait separately.
Monohybrid phenotypes:
3/4 round and 1/4 wrinkled 3/4 yellow and 1/4 green Dihybrid phenotypes (product rule):
Round and yellow: they are assorted independently R_Y_ = 3/4 X 3/4 = 9/16 Wrinkled and yellow: rrY_ = 1/4 X 3/4 = 3/16
Round and green: R_yy = 3/4 X 1/4 =3/16 Wrinkled and green: rryy = 1/4 X 1/4 = 1/16 Mendel’s Principle of Independent Assortment • Two hypotheses for gene assortment in a dihybrid cross: (b) Hypothesis: Independent Assortment (a) Hypothesis: Dependent assortment
P
Generation RRYY rryy RRYY Gametes RY ry (self) heterozygous for both Eggs
1/
2 1/
2 ry RY 1/
2 RY
1/
2 RrYy genes 1/
4 Eggs
Sperm 1/
4 ry F2
Generation 1/
4
1/
4 ry RRYY
RrYY RRYy
RrYy Sperm 1/
4 RrYY RrYy (self) RY rrYY rY
1/
4 RRYy Rryy rrYy
Rryy rryy Actual results
support hypothesis Ry
1/
4 RrYy RrYy
RRyy rrYy Actual results
contradict this hypothesis 1/
4 RY rY Ry dihybrid cross:
mating of parental
varieties differing
in 2 characteristics ry Gametes RY RrYy F1
Generation rryy ry 9/
16
3/
16
3/
16
1/
16 Yellow round
Green round
Yellow
wrinkled
Green
wrinkled Punnett square illustrating the genotypes
underlying a 9 : 3 : 3 : 1 ratio if you only look at the green color, how
many round and wrinkled do you have?
(the ratio). 3:1, because they are
independently assorted. same with yellow,
3:1 Dihybrid T Cross est
If the Independent Assortment Hypothesis is right, how many types
of gamete F1 plant RrYy can produce? In what ratio? The dihybrid should
make four types of
gametes in equal numbers RY rY Ry ry Test cross
Rr Yy round yellow F1 rr Yy wrinkled yellow X
Rr Yy
round yellow rr yy
wrinkled green you see 1:1:1:1 because they are independently
assorted and make the four types of gametes
which can be equally distributed 31
27 Rr yy round green 26 rr yy wrinkled green 26 Result → 1 : 1 : 1 : 1 ratio if the unknown parent is RrYy Summary One Trait (one gene)
3:1
1:1...
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 Spring '08
 Ribaya
 Zygosity

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