1. Genetics: Part II
Predicting Offspring

2. 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.

3. 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?

4. 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.
The “trick” to reading pedigrees is to look for a place where the parents are identical but at least one of their children are different from the parents. In this case, look at the parents in generation III. Both are un-shaded, meaning they do not show the trait in their phenotypes. BUT one child does show it. This means that the parents must have had the allele in question without showing it in their phenotype which means the trait must have been recessive (to “hide” in these two parents). Once that is known, the other questions can be answered. (The reverse situation can be seen in the previous slide in the second generation for widows peak)
First born male in generation II is on the left and must be Aa. With the same being true for the first female.

5. 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.
The chances of producing specific phenotypes can be calculated using the rules of probability. In order for us to determine the “chances” of an organism inheriting a single gene trait, we need to be able to apply the rules of probability.

6. 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

7. 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

8. 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.

9. 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.

10. 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)

11. 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

12. 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

13. 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

14. Complete the genetics card sort matching the term with its definition
Practice definitions
Genotype
Heterozygous
Phenotype
Allele
Monohybrid
Test cross
Dominant F1
Recessive F2 P1
Homozygous
Complete the genetics card sort matching the term with its definition

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

16. Created by:
Debra Richards
Coordinator of K-12 Science Programs
Bryan ISD
Bryan, TX