pre-ap biology tuesday february 6

Pre-AP Biology Tuesday February 6

Genetic Disorders and Solving Problems with Genetic Data Sets

If you were absent:

1. See the instructions on slides 3 – 4 for how to learn more about Cystic fibrosis and Huntington’s, two genetic disorders in humans

2. See slides 5 – 20 to practice working with phenotype ratio patterns based on monohybrid crosses

3. See slides 21 on for practice problem-solving using genetic data sets. Show your work (answers to the questions) in your packet at the top of P. 2

• http://teach.genetics.utah.edu/content/ngs/NGS-studentpacket.pdf

• Read P. 20 on Cystic Fibrosis

• Read P. 24 on Huntington’s

• See the Q. on the next slide for the most important information to get from the article (you do not need to write your answers down)

http://teach.genetics.utah.edu/content/ngs/NGS-studentpacket.pdf

Summarize genetic disorder:

1. Summarize symptoms that the person is expected to experience

2. Give the chromosome number and show the position of the affected gene on the chromosome

3. Describe the genetic mutation that results in the disease

4. Say what the inheritance pattern is

5. Tell how common the disease is

6. Summarize what the clinical outcome with screening and treatment is expected to be

7. Describe the genetic test for the disorder

Monohybrid Cross

Purple (P) is dominant to white (p)

Show phenotype ratios (purple: white)

a. Pp x Ppb. PP x ppc. PP x Ppd. Pp x pp

Dominant : Recessive

Options: 1 = 1:02 = 1:13 = 3:14 = 0:1

Monohybrid Cross

Purple (P) is dominant to white (p)

Show phenotype ratios (purple: white)

a. Pp x Pp 3:1b. PP x pp 1:0c. PP x Pp 1:0d. Pp x pp 1:1

Monohybrid Cross

Freckles (F) is dominant to nonfreckled (f)

Show phenotype ratios (freckled: nonfreckled)

a. Ff x ffb. FF x ffc. FF x Ffd. Ff x Ff

Dominant : Recessive

Options: 1 = 1:02 = 1:13 = 3:14 = 0:1

Monohybrid Cross

Freckles (F) is dominant to nonfreckled (f)

Show phenotype ratios (freckled: nonfreckled)

a. Ff x ff 1:1b. FF x ff 1:0c. FF x Ff 1:0d. Ff x Ff 3:1

Freckles is dominant. In a cross between two freckled parents, 9 children are freckled and 3 children are not.

1. Make a claim as to the genotypes of both parents.

2. Justify your claim by selecting an appropriate Punnett square from Packet P. 1:

In a cross between two freckled parents, 9 children are freckled and 3 children are not.

1. Make a claim as to the genotypes of both parents. Ff x Ff

2. Justify your claim using a Punnett square: F f

F

f

FF

Ff

Ff

ff

3 : 1

Freckles (F) is dominant to non-freckled (f). If a freckled person produces children with a non-freckled person in a ratio of 1:1, the parent’s genotypes are:

1. Not enough information given to tell2. FF and Ff3. FF and ff4. Ff and ff

For the following monohybrid cross, Tt x Tt (for T, dominant, versus dwarf, t, recessive) what would be the phenotypic ratio of the F1 generation tall:dwarf?

1. 1:02. 1:13. 3:24. 3:15. 1:2:1

For the following monohybrid cross, Tt x Tt(for T, dominant, versus dwarf, t, recessive) what would be the phenotypic ratio of the F1 generation tall:dwarf?

1. 1:02. 1:13. 3:24. 3:15. 1:2:1

Tall (T) is dominant to dwarf (t) in pea plants. If a tall pea plant is crossed with a dwarf pea plant and the phenotype ratio of the offspring is 1:1, the parents’ genotypes are:

1. TT and tt2. Tt and Tt3. Cannot be determined4. Tt and tt

In rabbits, black fur (B) is dominant to brown hair (b). What is the phenotype ratio (black : brown) of a cross between two heterozygous individuals?

1. 3:12. 1:13. 2:14. 1:3 5. 1:2:1

True-breeding = homozygous

AA ORaa

A true-breeding red-haired wuzzle is bred with a true-breeding blue-haired wuzzle. All the offspring are red-haired.

When these wuzzles are interbred, 75% are red-haired and 25% are blue-haired.

1. Make a claim with justification for which hair color is dominant:

2. What is the phenotype ratio for the first cross?3. Justify by selecting a Punnett square for this cross:4. What is the phenotype ratio for the second cross?5. Justify by selecting a Punnett square and

phenotype ratio for the first cross.

R R

r

r

Rr

Rr

Rr

Rr

R r

R

r

RR

Rr

Rr

rr

1:0 3:1

Cross 1: A purple-flowering plant is crossed with a blue-flowering plant, resulting in 247 purple-flowering and 262 blue-flowering plants.

Cross 2: Another purple-flowering plant is crossed with another blue-flowering plant, resulting in 460 purple plants and no blue-flowering plants

1. Which phenotype is dominant? Justify your selection:

2. Make a claim regarding the genotypes of the parents in cross 1. Justify your claim with a Punnett square.

3. Make a claim regarding the genotypes of the parents in cross 2. Justify your claim with a Punnett square.

Cross 1: A purple-flowering plant is crossed with a blue-flowering plant, resulting in 247 purple-flowering and 262 blue-flowering plants.Cross 2: Another purple-flowering plant is crossed with another blue-flowering plant, resulting in 460 purple plants and no blue-flowering plants

1. purple is the dominant because this is the only phenotype observed in cross 2.

2. Cross 1: Pp x pp would yield the observed 1:1 phenotype ratio.

3. Cross 2: PP x pp would yield the observed 1:0 phenotype ratio

P P

p

p

Pp

Pp

Pp

Pp

P p

p

p

Pp pp

ppPp


1. What phenotype ratio would result from a cross of two of the purple-plants that result from cross 1?

2. Justify your claim with a Punnett square:

P p

p

p

Pp pp

ppPp


1. What phenotype ratio would result from a cross of two of the purple-plants that result from cross 1? Pp x Pp would yield a 3:1 phenotype ratio

2. Justify your claim with a Punnett square

P p

p

p

Pp

Pp

pp

pp

P p

P

p

PP

Pp

Pp

pp

2 purple plants resulting from cross 1

Predicted 3:1 ratio

Cross 1: True-breeding bronze-eyed flies were crossed with true-breeding red-eyed flies. All of the F1 offspring had bronze-eyes.

Cross 2: F1 flies were next crossed resulting in 3,720 bronze-eyed and 1,260 red-eyed flies.

1. Which phenotype is dominant? Justify your selection:

2. Make a claim regarding the genotypes of the flies in the F1 generation.

3. Use a Punnett square to show how your answer to #2 above would yield the experimental results shown for cross 2:

Cross 1: True-breeding bronze-eyed flies were crossed with true-breeding red-eyed flies. All of the F1 offspring had bronze-eyes.

Cross 2: F1 flies were next crossed resulting in 3,720 bronze-eyed and 1,260 red-eyed flies.

1. Bronze-eyes are dominant because red-eyes are not seen (masked) in cross 1.

2. Parental: BB x bb resulting in F1: Bb

B b

B

b

BB

Bb

Bb

bb

3. Bb x Bb would yield a 3:1 ratio

pre-ap biology tuesday february 6

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