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Genetics 14.2.
 Monohybrid, Dihybrid, Probability rules, Independent assortment and meiosis.
Monohybrid – one mix.
Dihybrid – two mix.
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Independent assortment – identified by following TWO charcters at the SAME time.
Fig. 14.8 Seed color and seed shape.
Take true breeding pea plants.
P generation
YYRR x yyrr
YR
yr
Either alleles are passed on in the same pattern as parents received them or the two pairs of
alleles segregate independently so you see all possible combinations: YR, Yr, yR and yr.
F1 generation
YRyr Both Y and R expressed so hard to tell how inheritance working
so need to self-pollinate F1 to get F2. This is the key step to see how alleles being transmitted.
So, true breeding P generation give rise to F1 that self pollinate to give F2 where the ratios are
seen.
F2 generation: If passed on in same pattern then should get a 3:1.
F2 generation: If alleles segregate independently then should get a 9:3:3:1.
When Mendel did these crosses for all seven pea characteristics in different combinations he
ALWAYS got a 9:3:3:1 phenotypic ratio. Remember he could only quantify the phenotypes- what is
expressed.
Genotypic ratio 1:2:1:2:4:2:1:2:1.
 Only see this for genes on different chromosomes. If genes are on the same chromosome
and close together they are inherited together.
 ALL pea characteristics Mendel studied are located on separate chromosomes. This was
very lucky choice and made interpretation easier.
 Law of segregation applies to alleles of a single trait separating so each gamete gets one
of each allele.
 Law of independent assortment applies to two pairs of alleles for two different traits
separating independently of each other.
Multiplication rule – determines the probability of 2 or more independent events occurring together
in specific combinations. Give individual probabilities.
Ex. 2 coins tossed will come up heads? Multiply probability of one by the probability of the other.
Head coin 1 = 1/2
Head coin 2 = ½
So probability of both coming up heads is ½ x ½ = ¼.
For a monohybrid cross:
Rr x Rr
R
r
R
r
To get rr = ½ x ½ = ¼
To get RR = ½ x ½ = ¼.
For Heterozygotes use a second rule – addition rule.
Probability that any two or more events will occur is calculated by ADDING their INDIVIDUAL
probabilities.
So, one way to get Rr is R from egg and r from sperm = ¼
Other way to get Rr is r from egg and R from sperm = ¼
So, probability of F2 Rr = ¼ + ¼ = ½.
Use this method to solve complex genetic problems without use of Punnet Square.
Dihybrid corsses are equivalent to two or more independent monohybrid crosses occurring at the
SAME time so can apply rules of monohybrid cross to multiple traits.
Fig. 14.8 Ex. YyRr
Y = Yellow seed R = Round shape.
y = green seed r = wrinkled shape.
Monohybrid:
For YY = ½ x ½ = ¼.
For Yy = ¼ + ¼ = ½. (two independent events so addition rule).
For yy = ½ x ½ = ¼.
For RR = ½ x ½ = ¼.
For Rr = ¼ + ¼ = ½.
For rr = ½ x ½ = ¼.
Use these to calculate probabilities of F2 plant having:
YYRR = ¼ x ¼ = 1/16.
YyRR = ½ x ¼ = 1/8.
Pg. 260 Text.
1. CC x Cc What proportion off offspring Homo D? Homo r? Hetero?
2. BbDD x BBDd. What are each of the possible genotypes of the offspring and the
probability for each one?
3. What are the chances of offspring in 2 expressing either recessive trait?
Trihybrid cross:
P = Purple flower
P = white
Y = Yellow seed R = Round seed
y = green seed r = wrinkled seed
PpYyRr x Ppyyrr
What fraction of the offspring from this cross would exhibit recessive phenotypes for at least 2 of
the three characteristics?
a) List all the genotypes that fulfill conditions:
ppyyRr
ppyyrr
PPyyrr
Ppyyrr
ppYyrr
b) Now calculate probabilities of the genotypes by multiplying individual probabilities for allelic
pairs.
ppyyRr ¼ x ½ x ½ = 1/16.
Ppyyrr ½ x ½ x ½ = 1/8.=2/16.
PPyyrr ¼ x ½ x ½ = 1/16.
Ppyyrr ½ x ½ x ½ = 1/8 = 2/16.
ppYyrr ¼ x ½ x ½ = 1/16.
For hetero and homo from Yy x yy ½ x ½ = ¼ and ½ x ½ = ¼.
Then ¼ + ¼ = ½.
Now add all probabilities = 7/16.
Pg. 272 # 7 and 8.
7. AaBbCcDd x AaBbCcDd
aabbccdd ¼ x ¼ x ¼ x ¼ = 1/256.
AaBbCcDd ½ x ½ x ½ x ½ = 1/16.
AABBCCDD ¼ x ¼ x ¼ x ¼ = 1/256.
AaBBccDd ½ x ¼ x ¼ x ½ = 1/64.
AaBBCCdd ½ x ¼ x ¼ x ¼ = 1/128.
8.
AABBCC x aabbcc = AaBbCc
AABbCc x AaBbCc = AAbbCC
AaBbCc x AaBbCc = AaBbCc
aaBbCC x AABbcc = AaBbCc
1
½ x ¼ x ¼ = 1/32.
½ x ½ x ½ = 1/8.
1 x ½ x 1 = ½.