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Ch. 11 Notes - Genetics

Ch. 11 Notes - Genetics

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Ch. 11 Notes - Genetics. I. Who was Gregor Mendel?. A. Gregor Mendel - Austrian monk who studied how traits a re inherited ; known as the “ Father of Genetics ” 1. Genetics - branch of biology that studies heredity a. Heredity - passing on of traits from parent to offspring. - PowerPoint PPT Presentation

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Page 1: Ch. 11 Notes - Genetics

Ch. 11 Notes - Genetics

Page 2: Ch. 11 Notes - Genetics

A. Gregor Mendel - Austrian monk who studied how traits are inherited; known as the “Father of Genetics”

1. Genetics - branch of biology that studies heredity

a. Heredity - passing on of traits from parent to offspring

I. Who was Gregor Mendel?

Page 3: Ch. 11 Notes - Genetics

B. Mendel’s Experiments1. Studied/Researched on pea plants

a. They have many traits such as flower color (purple or white), peas (round or wrinkled), pea color (yellow or green) and height (tall or short).

I. Who was Gregor Mendel?

Page 4: Ch. 11 Notes - Genetics

b. Mendel bred a tall pea plant with a short plant (P generation).

c. The offspring in the 1st generation (F1

generation) were all tall.

I. Who was Gregor Mendel?

Short pea plant Tall pea plant

All tall pea plants

3 tall: 1 short

P1

F1

F2

Page 5: Ch. 11 Notes - Genetics

d. Then, he bred two of the F1 plants.

e. The offspring in the 2nd generation (F2 generation) were 75% tall and 25% short.

f. The short trait disappeared in the 1st generation and reappeared in the 2nd generation.

I. Who was Gregor Mendel?

Short pea plant

Tall pea plant

All tall pea plants

3 tall: 1 short

P1

F1

F2

Page 6: Ch. 11 Notes - Genetics

2. Mendel discovered that each trait is controlled by alleles. a. Alleles: forms/versions of the same gene

Each person has TWO alleles for each gene; 1 from mother and 1 from father.

Ex: T = the allele for tall; t = the allele for short

I. Who was Gregor Mendel?

Short pea plant

Tall pea plant

All tall pea plants

3 tall: 1 short

P1

F1

F2

Page 7: Ch. 11 Notes - Genetics

◦Some alleles are dominant and recessive Dominant - a trait that is always expressed

(seen) in an individual if the allele is present (capital letter) Ex: T = tall allele

Recessive - a trait that is hidden by the dominant allele; expressed only when two copies of the recessive allele are inherited (lowercase letter) Ex: t = short allele

I. Who was Gregor Mendel?

Page 8: Ch. 11 Notes - Genetics

◦Individuals can be described by their genotype and phenotype Genotype - genetic makeup (letters); alleles

present in an individual Ex: TT, Tt, tt

Type of Genotypes: Homozygous - two of the SAME alleles for a trait

Ex: TT or tt Heterozygous (hybrid) - two DIFFERENT alleles for a

trait Ex: Tt

I. Who was Gregor Mendel?

Page 9: Ch. 11 Notes - Genetics

◦Phenotype - physical appearance; what traits are expressed in the individual Ex: tall or short plants

I. Who was Gregor Mendel?

TT, Tt=tall tt=short

Page 10: Ch. 11 Notes - Genetics

A. Punnett Square/Test Cross - a tool used to predict the alleles/traits present in offspring

1. When parents produce gametes (sperm or eggs), their genes separate to produce

haploid cellsa. Each gamete contains ONE allele

II. Punnett Squares/Test Crosses

T T

Homozygous Tall DadHeterozygous Tall Dad

T t t t

Homozygous Short Dad

Page 11: Ch. 11 Notes - Genetics

B. How to Solve a Monohybrid Cross1. Determine the genotypes (letters) of the

parents. 2. Set up the punnett square with one parent

on top and one parent on the side.3. Fill out the Punnett square boxes (Look up

and over to the left to fill them in).4. Analyze the probability (likelihood) that

the offspring would possess each specific trait.

II. Punnett Squares/Test Crosses

Page 12: Ch. 11 Notes - Genetics

C. PracticeY = yellow pea color; y = green pea color

1. Cross two heterozygous pea plants. Parent 1 Genotype __________ Parent 2 Genotype __________ What is the chance of the offspring having yellow pea

color?

II. Punnett Squares/Test Crosses

Genotypic Ratio:

Phenotypic Ratio:

YY : Yy : yy

yellow : green

Page 13: Ch. 11 Notes - Genetics

Y = yellow pea color; y = green pea color2. Cross a homozygous recessive pea plant with

a heterozygous pea plant. Parent 1 Genotype __________ Parent 2 Genotype __________ What is the chance of the offspring having yellow pea

color?

II. Punnett Squares/Test Crosses

Genotypic Ratio:

Phenotypic Ratio:

YY : Yy : yy

yellow : green

Page 14: Ch. 11 Notes - Genetics

Y = yellow pea color; y = green pea color3. Cross a homozygous recessive pea plant with

a homozygous dominant pea plant. Parent 1 Genotype __________ Parent 2 Genotype __________ What is the chance of the offspring having yellow pea

color?

II. Punnett Squares/Test Crosses

Genotypic Ratio:

Phenotypic Ratio:

YY : Yy : yy

yellow : green

Page 15: Ch. 11 Notes - Genetics

D. How to Solve a Dihybrid Cross1. Determine parent genotypes (letters).

2. Determine gamete combinations from mom and dad. (FOIL method from math class - Use arrows!)3. Write the gametes from mom on 1 side of the

square and dad on the other side.4. Fill in the boxes in the punnett square (look up and to

the left).5. Analyze the probability (likelihood) that the offspring would possess the specific traits.

II. Punnett Squares/Test Crosses

Page 16: Ch. 11 Notes - Genetics

PRACTICE: R = round peas, r = wrinkled peas Y = yellow pea color; y = green pea color

◦Cross two heterozygous round yellow pea plants. Parent 1 Genotype _____ Parent 2 Genotype _____

Parent Gamete Possibilities: FOIL

Parent #1 Gametes: RY, Ry, rY, ryParent #2 Gametes: RY, Ry, rY, ry

II. Punnett Squares/Test Crosses

Page 17: Ch. 11 Notes - Genetics

What is the chance of the offspring having round and yellow peas?

RY

RY

Ry

Ry

rY

rY

ry

ry

Page 18: Ch. 11 Notes - Genetics

A. Autosomes - 22 pairs of body chromosomes in a human

III. Special Inheritance

22 Autosomes Sex chromosomes

Page 19: Ch. 11 Notes - Genetics

B. Sex chromosomes - 1 pair of chromosomes in a human; the last pair in a karyotype

a. Ex. XX-female XY-male

III. Special Inheritance

22 Autosomes Sex chromosomes

Page 20: Ch. 11 Notes - Genetics

Show a punnett square crossing male and female sex chromosomes.

III. Special Inheritance

d. Every time a male and female have a baby, what is the chance of them having a son? A daughter?

e. If a female has 5 sons in a row, what is the chance of her having another son? A daughter?

f. Which parent determines the sex of the child? WHY?

Page 21: Ch. 11 Notes - Genetics

C. Sex-Linked Traits - traits located on the sex chromosomes (usually the X chromosome)

a. Dad gives X chromosome to daughters and Y chromosome to sonsb. Mom gives X chromosome to daughters and sons

If a male inherits a sex-linked trait, which parent(s) gave him the trait? MOM

If a female inherits a sex-linked trait, which parent(s) gave her the trait? MOM OR DAD

Who (males or females) is most likely to inherit a sex-linked trait? WHY? Males – they only need to inherit ONE X chromosome

III. Special Inheritance

Page 22: Ch. 11 Notes - Genetics

c. Examples in humans Male pattern baldness Red/green colorblindness Hemophilia (problems with blood clotting) Muscular Dystrophy (muscle weakness, loss of

muscle tissue)

III. Special Inheritance

Page 23: Ch. 11 Notes - Genetics
Page 24: Ch. 11 Notes - Genetics

d. Other Examples Eye color in Drosophila (fruit flies)

R = Red eye Color; r = white eye color

Cross white eyed male (XrY) with red eyed female (XRXR)

III. Special Inheritance

What percentage of offspring are likely to have red eyes?

What percentage of female offspring are likely to have red eyes?

What percentage of male offspring are likely to have red eyes?

Page 25: Ch. 11 Notes - Genetics

◦ Examples Eye color in Drosophila (fruit flies)

R = Red eye Color; r = white eye color

Cross white eyed male (XrY) & heterozygous red eyed female (XRXr)

III. Special Inheritance

What percentage of offspring are likely to have red eyes?

What percentage of female offspring are likely to have red eyes?

What percentage of male offspring are likely to have red eyes?

Page 26: Ch. 11 Notes - Genetics

A. Incomplete Dominance - heterozygous individuals display an intermediate (blending) phenotype of the two homozygous individuals

1. Examples in humans Hair texture

SS = straight CC = curly SC = wavy

Tay Sachs Disease (inability to produce enzyme that breaks down lipids)

EE = produces enzyme NN = does not produce enzyme EN = produces half amount of enzyme

IV. Not all traits are controlled by simple dominant and recessive alleles!

Page 27: Ch. 11 Notes - Genetics

2. Other Examples Flower color in snapdragons

RR = Red, WW = white, RW = pink Cross a red flower with a white flower. Parent 1 Genotype ______ Parent 2 Genotype _____ What is the chance of producing a pink flower?

IV. Not all traits are controlled by simpledominant and recessive alleles!

Phenotypic Ratio: red: pink: white

Genotypic Ratio: RR : RW : WW

Page 28: Ch. 11 Notes - Genetics

Cross a pink flower with a pink flower. Parent 1 Genotype _____ Parent 2 Genotype _____ What is the chance of producing a white flower?

III. Not all traits are controlled by simple dominant and recessive alleles!

Phenotypic Ratio: red: pink: white

Genotypic Ratio: RR : RW : WW

Page 29: Ch. 11 Notes - Genetics

B. Co-dominance - heterozygous individuals display BOTH traits of the two homozygous individuals

1. Examples in humans Sickle Cell Anemia (abnormally shaped red blood cells)

NN = normal shaped cells SS = sickle shaped cells NS = normal and sickled shaped cells

Blood Type Type A Type B Type AB

III. Not all traits are controlled by simple dominant and recessive alleles!

Page 30: Ch. 11 Notes - Genetics

***** Special Inheritance *****◦ Hemoglobin—protein that carries oxygen in blood,

makes blood red In homozygous recessive individuals—hemoglobin is

defective and makes blood cells sickle (half moon) shaped These blood cells—cause slow blood flow, block small

vessels, tissue damage and pain In heterozygous individuals – both normal and sickled

hemoglobin are produced They produce enough normal hemoglobin that they do not

have serious health problems

Page 31: Ch. 11 Notes - Genetics
Page 32: Ch. 11 Notes - Genetics

2. Other Examples Coat color in chickens

BB = black WW = white BW = black AND white speckled

Cross a black rooster with a white chicken. Parent 1 Genotype _____ Parent 2 Genotype _____ What is the chance of producing black and white chicks?

IV. Not all traits are controlled by simple dominant and recessive alleles!

Phenotypic Ratio: black: black/white: white

Genotypic Ratio: BB: BW : WW

Page 33: Ch. 11 Notes - Genetics

IV. Not all traits are controlled by simple dominant and recessive alleles!

Complete the last cross on your own. Be ready to discuss.

Page 34: Ch. 11 Notes - Genetics

A. An example of multiple alleles – more than one allele controls the traitB. It is determined by the presence or absence of proteins (chains of amino acids) on the surface of red blood cells

a. Mixing incompatible blood types can cause blood clots, which can result in death

V. Blood Type (codominance in humans)

Page 35: Ch. 11 Notes - Genetics

Human Blood Types

V. Blood Type (codominance in humans)

Phenotype

Genotype Blood cell surface molecules (antigens)

Type A IAIA or IAi A antigensType B IBIB or IBi B antigensType AB IAIB A and B antigens

Type O ii No antigens

Page 36: Ch. 11 Notes - Genetics

V. Blood Type (codominance in humans)

Page 37: Ch. 11 Notes - Genetics

V. Blood Type (codominance in humans)C. Alleles IA and IB – are co-dominant to each other

D. Allele i –is recessive to both IA and IBa. Type O blood — universal donor Has no proteins on the blood cells so any blood type can

receive it

b. Type AB blood — universal acceptor

Has both proteins on blood cells so this blood type can receive any blood

Page 38: Ch. 11 Notes - Genetics

E. Cross parent with A ( IAi) blood with a parent with B blood (IBi).

V. Blood Type (codominance in humans)

IA i

IB

i

IAIB IBi

IAi ii

Genotypic ratio0 IA IA : 1 IA i : 0 IB IB :1 IB i : 1 IA IB : 1 ii

Phenotypic ratio (blood type)—1 type AB : 1 type A : 1 type B : 1 type O

Page 39: Ch. 11 Notes - Genetics

A. Rh Positive = have proteinsa. Genotypes: Rh+/Rh+ or Rh+/Rh-

B. Rh Negative = no proteinsb. Genotype: Rh-/Rh-

VI. Rh Factor – describe the presence or absence of another protein on blood cells.

Page 40: Ch. 11 Notes - Genetics

C. Cross parent heterozygous for the Rh factor with another parent who does not have the Rh factor.

Rh+ Rh-

Rh-

Rh-

Rh+/Rh- Rh-/Rh-

Genotypic ratio0 Rh+/Rh+ : 2 Rh+/Rh-: 2 Rh-/Rh-

Rh+/Rh-Rh-/Rh-

Phenotypic ratio2 Rh positive: 2 Rh negative

VI. Rh Factor – describe the presence or absence of another protein on blood cells.