Genetics: Part II
Predicting Offspring
2
Key
Male
1st
generation
Affected
male
Female
Affected
female
Mating
1st
generation
Ww
ww
Ww
ww
2nd
generation
Ww
ww
3rd
generation
WW
or
Ww
Widow’s
peak
ff
ff
(a) Is a widow’s peak a dominant or
recessive trait?
Ff
Ff
Ff
ff
ff
FF
or
Ff
3rd
generation
ww
No widow’s
peak
ff
Ff
2nd
generation
FF or Ff
Ww ww ww Ww
Ff
Offspring
Attached
earlobe
Free
earlobe
b) Is an attached earlobe a dominant
or recessive trait?
Interpret this Pedigree
Is the trait dominant or recessive?
What is the genotype of the first born male in generation II?
What is the genotype of the youngest female in generation II?
Explain.
Curriculum Framework
3A EK 3 inheritance provides an understanding
of the pattern of passage (transmission) of genes
from parent to offspring.
a. Rules of probability can be applied to
analyze passage of single gene traits from
parent to offspring.
Probability
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Rule of Addition
• Rule of addition: Chance that an event can
occur 2 or more different ways.
– Sum of separate probabilities
– Ex.1/4 Pp +1/4 Pp 1/2 Pp
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Rule of Multiplication
• The multiplication rule states that the
probability that two or more independent events
will occur together is the product of their
individual probabilities
• Chance that 2 or more independent events will
occur together
– Ex. Probability that 2 coins tossed at the same
time will land heads up
– Probability of H x H  HH
–½ x ½ = ¼
Rule of Multiplication
Cross: GgSs x GgSS
• What is the probability of producing green,
smooth seeds in this cross?
• Solution
• Green = 3/4
Smooth = 4/4
• 3/4 X 4/4 = 12/16 = 3/4 probability of
producing green smooth seed
From your formula chart:
If A and B are mutually exclusive,
then P (A or B) = P (A) + P (B)
If A and B are independent,
then P (A and B) = P(A) X P(B)
Ex. Probability of a couple having three girls?
Ex. Probability of a couple having three boys?
Ex. Probability of having three boys or three girls?
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For example:
In a heterozygous cross YyRr
Probability of YYRR  1/4 (probability of YY)  1/4 (RR)  1/16
Probability of YyRR  1/2 (Yy)
 1/4 (RR)  1/8
Cross PpYyRr x PPyyrr
ppyyRr
ppYyrr
Ppyyrr
PPyyrr
ppyyrr
1/
1/ (yy)  1/ (Rr)
(probability
of
pp)

 1/16
4
2
2
 1/16
?
 ?
 1/16
Chance of at least two recessive traits
 6/16 or 3/8
Cross PpYyRr x Ppyyrr (Answer)
ppyyRr
ppYyrr
Ppyyrr
PPyyrr
ppyyrr
1/ (yy)  1/ (Rr)
(probability
of
pp)

4
2
2
1/  1/  1/
4
2
2
1/  1/  1/
2
2
2
1/  1/  1/
4
2
2
1/  1/  1/
4
2
2
1/
Chance of at least two recessive traits
 1/16
 1/16
 2/16
 1/16
 1/16
 6/16 or 3/8
Practice definitions
•
•
•
•
•
•
•
Genotype
Phenotype
Monohybrid
Dominant
Recessive
P1
Homozygous
•
•
•
•
•
Heterozygous
Allele
Test cross
F1
F2
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Practice
• Now that you have reviewed some basic
genetics concepts solidify your skill by
completing the set of practice problems
available at
http://anthro.palomar.edu/practice/mendqui2.htm
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Created by:
Debra Richards
Coordinator of K-12 Science Programs
Bryan ISD
Bryan, TX