1. Agenda 4/5 Genetics Intro Review Predicting Offspring Lecture
Pedigree Practice
Homework: Punnett Square video and notes, Chp 11 Notes
Turn in: Chp 11 notes if finished
TEST MOVED TO WEDNESDAY 4/13
*Place phones in your folder for the duration of class*

2. Give an example of a heterozygous genotype and a homozygous genotype

3. Genetics: Part II
Predicting Offspring

4. The answer, of course, is no
The answer, of course, is no. However, this is a common misconception or misunderstanding about how the numbers work in inheritance. Today we will look at a couple of tools to use, in addition to a Punnett square, to study and predict offspring of specific crosses.

5. Is a widow’s peak a dominant or recessive trait? b)
Key
Male
Female
Affected male
Affected female
Mating
Offspring
1st generation Ff Ff ff Ff
1st generation Ww ww ww Ww
2nd generation
2nd generation
FF or Ff ff ff Ff Ff ff Ww ww ww Ww Ww ww
3rd generation
3rd generation ff
FF or Ff
WW or Ww ww
A valuable tool in studying inheritance patterns is a pedigree. A pedigree is a family tree that describes the interrelationships of parents and children across generations
Inheritance patterns of particular traits can be traced and described using pedigrees
Widow’s peak
No widow’s peak
Attached earlobe
Free earlobe
(a)
Is a widow’s peak a dominant or recessive trait? b)
Is an attached earlobe a dominant or recessive trait?

6. Pedigree Symbols

7. Pedigree #1

8. Pedigree #1

9. Pedigree #1

10. Pedigree #1 Characteristics of a dominant pedigree:
1) Every affected individual has at least one affected parent
2) Affected individuals who mate with unaffected individuals have a 50% chance of transmitting the trait to each child; and
3) Two affected individuals may have unaffected children.

11. Pedigree #2

12. Pedigree #2

13. Pedigree #2

14. Pedigree #2 Characteristics of recessive pedigrees:
1) An individual who is affected may have parents who are not affected
2) All the children of two affected individuals are affected
3) In pedigrees involving rare traits, the unaffected parents of an affected individual may be related to each other.

15. Pedigree #3

16. Pedigree #3

17. Pedigree #3

18. Pedigree #3 Characteristics of sex-linked pedigrees
All females from the affected male have the trait, but not all males from affected females (dominant)
All males from the affected female have the trait, but not all females from affected males (recessive)

19. Label the genotypes and identify type of inheritance

20. Probability
Mendel’s laws of segregation and independent assortment reflect the rules of probability
When tossing a coin, the outcome of one toss has no impact on the outcome of the next toss
In the same way, the alleles of one gene segregate into gametes independently of another gene’s alleles

21. 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
The addition rule states that the probability that any one of two or more exclusive events will occur is calculated by adding together their individual probabilities
The rule of addition can be used to figure out the probability that an F2 plant from a monohybrid cross will be heterozygous rather than homozygous

22. 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
The multiplication rule (also called the product rule) states that the probability that two or more independent events will occur together is the product of their individual probabilities.
Probability in an F1 monohybrid cross can be determined using the multiplication rule.
Segregation in a heterozygous plant is like flipping a coin: Each gamete has a 1/2 chance of carrying the dominant allele and a 1/2 chance of carrying the recessive allele.

23. Rule of Multiplication
Cross: GgSs x GgSS
What is the probability of producing green, smooth seeds in this cross?
Solution
Green = 3/ Smooth = 4/4
3/4 X 4/4 = 12/16 = 3/4 probability of producing green smooth seed
Using the multiplication rule can expedite the process of predicting the probability of producing specific phenotypes in the offspring. For example, in peas, green seed color (G) is dominant to yellow seed color (g). Smooth seed coats (S) are dominant to wrinkled seed coats (s). Given that, determine the probability of producing green, smooth seeds in the cross shown here.

24. 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?
The AP Biology Formula chart contains the formula to use when applying the probability rule of addition.
The probability of having one girl is ½. The probability of having a second girl is ½. The probability of having the third girl is ½. Each conception is independent of the next. So… ½ x ½ x ½ = 1/8 chance of having three girls. The same would be true of having three boys.
However, the odds of producing EITHER three boys or three girls is ¼ because there are two ways to get the result. So the probabilities are added (1/8 + 1/8 = ¼)

25. In a heterozygous cross YyRr
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
We can apply the multiplication and addition rules to predict the outcome of crosses involving multiple characters
A dihybrid or other multicharacter cross is equivalent to two or more independent monohybrid crosses occurring simultaneously
In calculating the chances for various genotypes, each character is considered separately, and then the individual probabilities are multiplied

26. Cross PpYyRr x Ppyyrr ppyyRr
1/4 (probability of pp)  1/2 (yy)  1/2 (Rr)
 1/16
ppYyrr
 1/16
Ppyyrr
 ?
PPyyrr
 ?
ppyyrr
 1/16
Chance of at least two recessive traits
When considering more than two traits, calculating the probability mathematically is much simpler than drawing a punnett square. Work with a partner to write out how each probability should be calculated?
 6/16 or 3/8

27. Cross PpYyRr x Ppyyrr (Answer)
1/4 (probability of pp)  1/2 (yy)  1/2 (Rr)
 1/16
ppYyrr
1/4  1/2  1/2
 1/16
Ppyyrr
1/2  1/2  1/2
 2/16
PPyyrr
1/4  1/2  1/2
 1/16
ppyyrr
1/4  1/2  1/2
 1/16
Chance of at least two recessive traits
How do your answers compare?
 6/16 or 3/8

28. Exit Ticket Cross GgRRyyBb x GgRrYyBb
What is the probability you will get the genotype GGRRyyBb?

29. Practice
Now that you have reviewed some basic genetics concepts solidify your skill by completing the set of practice problems available at