1. Genetics

2. Check your Notes Review: Describe the following terms:
Gene:
is a unity of heredity made up of a section of DNA. One gene codes for one character.
Genotype:
The genotype of a particular individual is the combination of alleles that individual has for a particular gene.
Phenotype:
The phenotype of an individual is the physical expression of that individual’s genotype.
A genotype that contains the dominant allele will show the dominant phenotype.
A genotype that contains only recessive alleles will show the recessive phenotype.
Allele:
While only one section of DNA comprises a gene, that one section may have several different versions. These different versions are called alleles.

3. Review Continue: Dominant:
A dominant allele is one that is always expressed.
Dominant alleles are represented as capital letters.
Recessive:
A recessive allele is one that is not expressed if the dominant allele is also present.
Recessive alleles are represented as lowercase letters.
Homozygous:
A genotype that has two dominant alleles is said to be homozygous dominant. (RR)
A genotype that has two recessive alleles is said to be homozygous recessive. (rr)
Heterozygous:
A genotype that has one dominant and one recessive allele is said to be heterozygous. (Rr)

4. Mendelian Genetics
Vocabulary:
heredity
generation
character
parental (P) generation
trait
filial (F) generation
Heredity is the passing of genetic traits from parent to offspring.
How does this happen?

5. Gregor Mendel
19th century Austrian monk. b d Interested in inheritance patterns and studied pea plants in attempt to figure them out. Discovered basic patterns that are now the foundation of genetics – known as the Father of Genetics. Work went largely unrecognized until several decades after his death. Even Charles Darwin, who was working at the same time, never knew what Mendel had discovered.

6. They were easy to grow and count.
Why pea plants?
They were easy to grow and count.
Mendel could carefully select parents by using a brush to pollinate them.
They had traits that were easily identifiable.
*Distribute Biology Inquiries Worksheet 4.1

7. Tall x Short
First Mendel made sure he had “true breeding” plants. Then he crossed a true breeding tall plant to a true breeding short plant.
P (parental) generation
F1 (filial) generation X

8. First Filial Cross: Tall x Tall
Mendel crossed two tall plants from his F1 generation. X

9. (Show In Decimal Form Not Percentage) .5 = Yes / 50% = No
Punnett Squares
We use a grid system, known as the Punnett square, to help us solve genetics problems.
Given the genotypes of parents, we can predict the probability (likelihood expressed as a percentage or decimal) of the genotype of any particular offspring.
probability =
(Show In Decimal Form Not Percentage) .5 = Yes / 50% = No
number of one kind of possible outcome
total number of possible outcomes

10. Characters and Traits
The physical features that are inherited are called characters. The possible versions of those characters are called traits.
Characters?
flower color
height
coat color
eye color
Traits?
purple
tall
blue
gray

11. Generations
A generation is a group of offspring and decedents from a given group of parents.
P generation = Parental Generation
F1 generation = offspring of first cross
F2 generation = offspring of second cross
and so on…
P = Parental
F = Filial = Offspring as in “Fille” or “Fils”, French for Daughter and Son.

12. Which letter to use?
Does not matter in Genetics which letter you decide to use.
DO NOT have similar Capital letters and lower case letters.
Great = Aa, Bb, Dd, Ee, Ff, Gg, Hh, Nn, Qq, Rr, and Tt
Ok = Ii, Jj, Ll,
Bad = Cc, Kk, Mm, Oo, Pp, Ss, Uu, Vv, Ww, Xx, Yy, and Zz

13. Using a Punnett Square to correctly answer a word problem.
STEPS: 1. Read the entire problem first. 2. Make a key with the given facts a. Dominate/Recessive 3. Chose a letter to represent your facts 4. Write down your “Cross” (Mating) 5. Draw a Punnet Square Parent genotypes: TT and t t Cross T T  t t

14. Punnett square T T t T t T t T t T t T T  t t
6. "split" the letters of the genotype for each parent & put them "outside" the p-square
7. determine the possible genotypes of the offspring by filling in the p-square
8. summarize results (genotypes & phenotypes of offspring)
P generation
T T  t t
T T
F1 generation
Genotypes:
100% T t t
T t
T t
T t
T t
Phenotypes:
100% Tall plants

15. Check your Notes: Gregor Mendel Understand and use the terms:
Father of Genetics
Plants used
What he found
Understand and use the terms:
heredity
characters
traits
generation
P generation, F generations
allele
dominant
recessive
phenotype – dominant and recessive
genotype – dominant, recessive, heterozygous

16. Practice Problems
Find the Phenotype Ratio for the F2 Generation for all 7 Characters. Than indicate based on your evidence what the genotype must have been for the F1 Generation and explain your answer.
Character
Dominant Trait
Recessive Trait
F2-Generation
F2 Ratio
Seed Color
Yellow
Green
6,022 Yellow
2,001 Green
Seed Shape
Round
Wrinkled
5,474 Round
1,850 Wrinkled
Pod Color
428 Green
152 Yellow
Pod Shape
Inflated
Constricted
882 Inflated
299 Constricted
Plant Height
Tall
Short (Dwarf)
787 Tall
277 Short
Flower Color
Purple
White
705 Purple
224 White
Flower Position
Axial (Middle of branch)
Terminal (End of Branch)
651 Axial
207 Terminal

17. 1. For each genotype below, indicate whether it is a heterozygous OR homozygous.
TT _____ Pp _____ dd _____ Ff _____ Tt _____ FF _____
Which of the genotypes listed above would be considered purebred? _____________________

18. Answer
Homozygous - TT, dd, FF; Heterozygous- Pp, Ff, Tt; Purebred = TT, dd, FF

19. 2. In the Smurf family, a blue body color (B) is dominant to green (b)
2. In the Smurf family, a blue body color (B) is dominant to green (b). Determine the phenotype for each genotype below based on this information.
BB _________________
Bb _________________
bb _________________

20. BB - blue Bb - blue bb - green
Answer
BB - blue Bb - blue bb - green

21. 3. If tall eyeballs (T) are dominant to short eyeballs(t), give the genotypes that are possible for members of Mr. Leslers’ family.
Tall eyeballs = _________________
Short eyeballs = __________________

22. Tall eyeballs - TT, Tt Short eyeballs - tt
Answer
Tall eyeballs - TT, Tt Short eyeballs - tt

23. 4. Mr. Z is known for his big round eyes, which is dominant over an oval eye shape. If he is heterozygous for his round eye shape and marries a woman with oval eye shape, what type of eyes might the kids have?
A. List the genotypes for each:
Heterozygous round eyes - _______ Oval eyes - _______
B. Complete the Punnett square to show the possibilities that would result if Mr. Z had children with an oval-eyed woman.
C. List the possible genotypes and phenotypes for their children.
D. What are the chances of a child with a round eye shape? ____%
E. What are the chances of a child with an oval eye shape? ____%

24. Answer
A. Heterozygous round = Rr, Oval = rr
B. See square at right
C. Rr - round & rr - oval
D. 50%
E. 50%
R r r
R r
r r
R r
r r

25. 5. Patrick recently married Patti, a cute girl he met at a local dance
5. Patrick recently married Patti, a cute girl he met at a local dance. He is considered a purebred for his tall head shape (T), which is dominant over a short head (t). If Patti is a short- headed woman, what type of heads would their children have?
A. List the genotypes for each: Patrick - ______ Patti - _______
B. Complete the Punnett square to show the possible offspring.
C. Which type of head is most likely: tall or short? Explain.
D. Would the children be considered purebreds? Explain.

26. T T t T t T t T t T t Answer T T  t t A. Patrick - TT, Patti = tt
B. See square at right
C. Tall head is most likely, since all genotypes that result would represent a tall head (100%).
D. The children would not be considered purebreds, since they would each have a dominant gene and a recessive gene.
T T  t t
T T t
T t
T t
T t
T t

27. 6. Punnett squares are used by geneticists to determine the probability of different offspring genotypes. In the one shown below, what letter(s) belong in the lower right box?
a) Aa
b) AA
c) aa
d) a

28. Answer
aa - CORRECT --> All you have to do is carry over the letters in the row and column headings. The 2 empty boxes on the left will be Aa and the 2 on the right will be aa. In other words, 50% of the offspring are homozygous recessive (aa) and 50% are heterozygous (Aa).

29. 7. If two people who are both carriers for a genetically inherited fatal recessive disease decide to become parents, what will be the odds that their children will also be carriers?
a) 1 out of 4
b) 2 out of 4
c) 3 out of 4
d) 4 out of 4

30. Answer
2 out of 4 - CORRECT --> There will be a 50% chance (2 out of 4) of having a carrier (Aa) child. Similarly, each time there will be a 25% chance of having an offspring who will inherit the disease (aa) and die.

31. 8. If a woman is homozygous normal and her husband is heterozygous for a genetically inherited recessive disease and they decide to become parents, what is the probability that they will have a healthy child?
a) 1 out of 4
b) 2 out of 4
c) 3 out of 4
d) 4 out of 4

32. Answer
4 out of 4 - CORRECT --> All of the children will be healthy since none of them can be homozygous recessive (aa). However, there will be a 50% chance at each birth that the children will be carriers (Aa). The remaining 50% will be homozygous dominant (AA).

33. 9. If two parents are heterozygous for a genetically inherited dominant trait, what is the probability that they will have a child together who has this trait in his or her phenotype?
a) 25%
b) 50%
c) 75%
d) 100%

34. Answer
75% - CORRECT --> There is a 50% chance that the children will be heterozygous (Aa) and a 25% chance that they will be homozygous dominant (AA). Children with either of these genotypes will have this trait expressed in their phenotypes.

35. 10. If two parents are homozygous for a genetically inherited recessive trait, what is the probability that they will have a child who does not have this trait in his or her phenotype?
a) 0%
b) 25%
c) 50%
d) 100%

36. Answer
0% - CORRECT --> Since both parents are homozygous recessive (aa), all of their children will be as well. As a result, none of them will be free of this recessive trait in their phenotypes.

37. Probablility Vocabulary: Empirical Probability The Product rule
Theoretical Probability
The Sum Rule

38. Empirical Probability
An event is calculated by counting the number of times that event occurs and dividing it by the total number of times the event could have occurred.
Example Problem
The event you were looking for was a wrinkled pea seed, and saw it 1,850 times out of 7,324 total seeds you examined, the empirical probability of getting a wrinkled seed would be?
Answer
1850/7,324= or very close to 1 in 4 seeds.

39. Theoretical Probability
An event calculated based on information about the rules and circumstances that produce the event.
It reflects the number of times an event is expected to occur relative to the number of times it could possibly occur.
Example Problem
If you had a pea plant heterozygous for a seed shape gene (Rr) and let it self-fertilize, how many off spring would be homozygous recessive?
Answer
You could use the rules of probability and your knowledge of genetics to predict the outcome that 1 out of 4 off spring would be homozygous recessive (rr) and appear wrinkled, corresponding to a probability.

40. Rules of Live By:
Punnet Squares will not accurately predict the exact numbers of each genotype a cross will produce over the course of many offspring.
Punnett squares only tells you the probability of genotypes for each separate individual produced by a particular cross.
The larger the number of data points that are used to calculate an empirical probability, such as shapes of individual pea seeds, the more closely it will approach the theoretical probability.

41. The Product Rule
Definition - the probability of two independent events occurring together can be calculated by multiplying the individual probabilities of each event occurring alone
The Product Rule of probability is used to determine the probability of having both dominant traits in the F2 progeny; it is the product of the probabilities of having the dominant trait for each characteristic.

42. The Product Rule - Continue
Imagine that you are rolling a six-sided die (D) and flipping a penny (P) at the same time. The die may roll any number from 1–6 (D#), whereas the penny may turn up heads (PH) or tails (PT). The probability that you will obtain the combined outcome 2 and heads is?

43. Practice Problems
Two Heterozygous tall parents had 3 children, what is the probability that all 3 children are tall?
(.75)x(.75)x(.75)=.42 or 42% chance
Two Heterozygous tall parents had 3 children, what is the probability that all 3 children are short?
(.25)x(.25)x(.25)=.02 or 2% chance
Two Heterozygous tall parents had 3 children, what is the probability that all 2 children are short and 1 is tall?
(.25)x(.25)x(.75)=.05 or 5% chance
Two Heterozygous tall parents had 3 children, what is the probability that all 2 children are tall and 1 is short?
(.75)x(.75)x(.25)=.14 or 14% chance

44. If two people who are both carriers for a genetically inherited fatal recessive disease decide to become parents and have 5 children, what is the probability that all 5 children are carriers?
(.5)x(.5)x(.5)x(.5)x(.5)=0.031 or 3% chance
If two people who are both carriers for a genetically inherited fatal recessive disease decide to become parents and have 5 children, what is the probability that 3 children are carriers, 1 child inherited the disease and 1 child is not a carrier and healthy?
(.5)x(.5)x(.5)x(.25)x(.25)=0.01 or 1% chance
If two people who are both carriers for a genetically inherited fatal recessive disease decide to become parents and have 5 children, what is the probability that 3 children are not carriers and healthy and 2 children are carriers?
(.25)x(.25)x(.25)x(.5)x(.5)=0.004 or 0.4% chance
If two people who are both carriers for a genetically inherited fatal recessive disease decide to become parents and have 5 children, what is the probability that 3 children inherited the fatal disease and 2 are not carriers and healthy?
(.25)x(.25)x(.25)x(.25)x(.25)= or .09% chance

45. Mr. Smith had a male basset hound Droopy
Mr. Smith had a male basset hound Droopy. Droopy was well known for his big long ears, which is dominant over little short ears. Droopy is heterozygous for his big long ears and mates with Foxy who is also heterozygous. Foxy delivers 10 puppies, 7 males and 3 females. What is the probability that 5 puppies are long ears and 5 are short ears?
(.75)(.75)(.75)(.75)(.75)x(.25)(.25)(.25)(.25)(.25) =
(.75)5 x (.25)5 =
Mr. Smith had a male basset hound Droopy. Droopy was well known for his big long ears, which is dominant over little short ears. Droopy is heterozygous for his big long ears and mates with Foxy who is also heterozygous. Foxy delivers 10 puppies, 7 males and 3 females. What is the probability that 3 puppies are long ears and 7 are short ears?
(.75)3 x (.25)7 =

46. Mr. Smith had a male basset hound Droopy
Mr. Smith had a male basset hound Droopy. Droopy was well known for his big long ears, which is dominant over little short ears. Droopy is heterozygous for his big long ears and mates with Foxy who is also heterozygous. Foxy delivers 10 puppies, 7 males and 3 females. What is the probability that all 7 male puppies are long ears and the 3 females are short ears?
(.75)7 x (.25)3 x (.5)10 =
Mr. Smith had a male basset hound Droopy. Droopy was well known for his big long ears, which is dominant over little short ears. Droopy is heterozygous for his big long ears and mates with Foxy who is also heterozygous. Foxy delivers 10 puppies, 7 males and 3 females. What is the probability that 5 of the males have long ears. 2 - males have short ears. 2 – Females have short ears and 1-female has long ears?
(.5)5 x (.75)5 x (.5)2 x (.25)2 x (.5)2 x (.25)2 x (.5) x (.75) =