Unit 4: Cell Division & Heredity

Part 3

Ch. 14 & Ch. 15

Mendelian Genetics & Chromosomal Basis of Inheritance

I. Genetics As A Science

A. Genetics = Study of heredity

1. Heredity = How information gets transferred to us from our parents.

II. Genetics Vocabulary

A. Trait: Characteristic that is different among individuals.

Eye color, height, skin color

II. Genetics Vocabulary

B. Cross: When 2 individuals mate.

2 trees, 2 dogs, etc.

II. Genetics VocabularyC. Hybrid: Result of crossing individuals with

different traits.

III. Discovery of Genetics

A. Gregor Mendel: Used pea plants to explore heredity.

III. Discovery of Genetics

B. Mendel concluded 2 things1. “Factors” determine what organisms will be

like .

a. Genes – determine traits1. Height

2. Hair color

III. Discovery of Genetics

b. Each “factor” can come in different varieties.

1. Alleles- varieties of genes

a. Hair color gene

1. Brown allele

2. Blonde allele

3. Red allele

III. Discovery of Genetics

B. Mendel concluded 2 things…

2. Some alleles are “stronger” than others

III. Discovery of Genetics

a. Law of Dominance: Some alleles are dominant & others are recessive.

1. Dominant: Will always be seen.

2. Recessive: Only seen if dominant is not there.

IV. Mendel’s Work

A. Crossed a tall plant with a short plant.

1. P Generation = Original pair of plants.

(Parents)

IV. Mendel’s Work

B. Results1. All tall plants

F1 = First set of offspring.

(children)

IV. Mendel’s Work

C. Curious about why none were short.

1. Crossed 2 plants from the F1. generation.

Yes, mated 2 siblings!

IV. Mendel’s Work

D. Results

1. 75% Tall

25% Short

F2 = offspring of the F1 generation.

(grandchildren)

V. Explaining the Outcomes

A. The allele for “shortness” didn’t disappear.

B. The allele for “tallness” was dominant.

How was the recessive allele able to be “recovered”?

V. Explaining the Outcomes

C. Segregation: Alleles separate from each other during meiosis.

D. Gametes then combine differently during fertilization.

T t T t

T t T t

T t T t

T T t t

VIII. More About Alleles

A. Allele combinations

1. Have 2 alleles for each gene

Height – Tall or short

VIII. More About Alleles

B. Each one represented by the first letter of the dominant allele:

1. Dominant = Uppercase

2. Recessive = Lowercase

a. Height

T = Tall (Dominant)

t = Short (Recessive)

VIII. More About Alleles

a. If 2 of the SAME allele: HOMOZYGOUS

TT = Tall tt = Short

Homozygous Homozygous

Dominant Recessive

VIII. More About Alleles

b. If 2 DIFFERENT alleles: HETEROZYGOUS

Law of dominance: If dominant is present, it will “cover” the recessive.

T t = Tall

IX. Looks Can Be Deceiving!

A. Phenotype: Physical appearance (What’s on the outside)White, Red

B. Genotype: Genetic makeup (What’s on the inside) RR, Rr, rr

X. Punnett Squares

A. Diagrams for predicting results of any cross.

Genotype of parent 1

Genotypes of

offspring

Genotype of

parent 2

XI. Single Trait Cross

A. A cross for 1 trait.

1. Hair color ONLY

XI. Single Trait Cross

Genotype of parent 1

T T

Genotype of

parent 2

t t

T T

t

t

XI. Single Trait Cross

Results: All 4 combinations are the same.

Results vary depending on parents!

T T

t

t

T t

T t T t

T t

XI. Monohybrid Cross

F1 Cross: Cross 2 of the “kids”.

Results: 1:2:1 Genotypic Ratio

3:1 Phenotypic Ratio

T t

T

t

T T

T t t t

T t

XI. Monohybrid Cross

• RULE: According to the Mendel’s Laws, A monohybrid cross of F1 will ALWAYS result in 3:1 p ratio and 1:2:1 g ratio.

• If not… something else is acting.- Alternate types of dominance

XII. Dominant/Recessive??A. Simple rules of

dominance/recessive don’t always hold true.

1. Incomplete dominance:

Both alleles present = Intermediate

XII. Dominant/Recessive??

2. Codominance:

Both alleles present = Both appear

Brown & White

both dominant =

both colors

appear

XII. Dominant/Recessive??3. Multiple alleles:

More than 2 alleles for 1

gene

Blood Groups

XII. Dominant/Recessive??4. Lethal Dominant Traits

Monohybrid cross will yield a 2:1 ratio of dominant to recessive.

H.D. not viable so does not count in the total offspring. T t

T

t

T T

T t t t

T t

Sex Chromosomes

2. 44 Autosomes & 2 Sex chromosomes

a. Male: 46,XY b. Female: 46,XX

Sex Chromosomes

3. Sperm = X or Y 4. Egg = X only

Y

X

X X

XX XX

XY XY

Male

Female

50% female

50% male

Sex-Linked Genes

1. Found on the sex chromosomes, X or Y.

2. Y-linked traits = “maleness” mostly

3. X-linked traits = many traits- Colorblindness

Sex-Linked Genes

4. Males = Y Xa. All X-linked traits will be expressed in males.

Only 1 X chromosome, so either YES or NO!

5. Females = X Xa. If trait is dominant, it will be expressed if 1 copy is

present.

b. If trait is recessive, it will be expressed if present on BOTH X chromosomes.

More on X Chromosomes

1. Barr body: One of the X chromosomes in females “turns off” during early development.

2. Only in females; Males need the X!!

3. No baby ever born without an X

XIII. Two Trait Cross

A. A cross that looks at 2 traits.

1. R = right handed r = left handed

2. D = Dimples d = no dimples

The Testcross

• A person has brown hair, and brown hair is dominant. How can we tell the genotype?

• Testcross: crossing an unknown genotype with a homozygous recessive.– h.r. will always have KNOWN genotype.

» gg jj kk ee

The Testcross

• If homozygous dominant:– BB x bb = ALL Bb (all brown hair)

• If heterozygous:– Bb x bb = HALF Bb, Half bb (half brown, half

blonde)

XII. Dihybrid Cross

Genotype of parent 1 = R r D dGenotype of parent 2 = R r D d

1. Figure out the combinations for each parents’ gametes.

XII. Dihybrid Cross

Genotype of parent 1 = R r D dFIRST

R r X D d

R D

XII. Dihybrid Cross

Genotype of parent 1 = R r D dOUTSIDER r X D d

R d

XII. Dihybrid Cross

Genotype of parent 1 = R r D dINSIDER r X D d

r D

XII. Dihybrid Cross

Genotype of parent 1 = R r D dLASTR r X D d

r d

XII. Dihybrid Cross

Genotype of parent 1 = R r D d

Gamete combinations =

R D R d r D r d

XII. Dihybrid Cross

Genotype of parent 2 = R r D d

Same genotype, same

gamete combinations =

R D R d r D r d

XII. Dihybrid Cross2. Make a Punnett square: 16 possibilities

XII. Dihybrid Cross

3. Parents’ possible gametes

R D R d r D r d

R D

R d

r D

r d

XII. Dihybrid Cross

4. Perform Crosses

R D R d r D r d

R D

R d

r D

r d

RRDD

RRDd

RRDd

RRdd

RrDD

RrDd

RrDd

RrDD

RrDd

rrDD

Rrdd

RrDd

Rrdd

rrDd

rrDd rrdd

XII. Dihybrid Cross5. Results: 9 = right handed w/dimples

R D R d r D r d

R D

R d

r D

r d

RRDD

RRDd

RRDd

RRdd

RrDD

RrDd

RrDd

RrDD

RrDd

rrDD

Rrdd

RrDd

Rrdd

rrDd

rrDd rrdd

XII. Dihybrid Cross5. Results: 3 = Right handed, no dimples

R D R d r D r d

R D

R d

r D

r d

RRDD

RRDd

RRDd

RRdd

RrDD

RrDd

RrDd

RrDD

RrDd

rrDD

Rrdd

RrDd

Rrdd

rrDd

rrDd rrdd

XII. Dihybrid Cross5. Results: 3 = Left handed w/dimples

R D R d r D r d

R D

R d

r D

r d

RRDD

RRDd

RRDd

RRdd

RrDD

RrDd

RrDd

RrDD

RrDd

rrDD

Rrdd

RrDd

Rrdd

rrDd

rrDd rrdd

XII. Dihybrid Cross5. Results: 1 = Left handed, no dimples

R D R d r D r d

R D

R d

r D

r d

RRDD

RRDd

RRDd

RRdd

RrDD

RrDd

RrDd

RrDD

RrDd

rrDD

Rrdd

RrDd

Rrdd

rrDd

rrDd rrdd

XII. Dihybrid Cross5. Results:

The probability of having a child with dimples and being left-handed?

3/16 = r r D D r r D d r r d DOut of 16, 3 should be left handed and have dimples.

XII. Dihybrid Cross

6. Independent Assortment

a. New combo’s were made

b. Traits didn’t affect each other

c. Gene for right hand/left hand independent from gene for dimples… why?

XII. Dihybrid Cross

6. Independent Assortment

d. On different chromosomes!

* Different genes can assort without affecting each other if they are on different chromosomes.

XII. Dihybrid Cross

Phenotypic Ratio is always: 9:3:3:1

If not… something else is going on… Alternate type of dominance, sex linked trait…

Probability

A. Punnett squares can get too large if we look at more than 2 traits. Using rules of probability are much easier!

1. Rule of multiplicationa. signaled by the word “AND”- what’s the probability I will have a

boy and he will have blue eyes?

Probability

1. Rule of multiplication boy = XY girl = XX blue eyes = ggGgXX (mom) x GgXY (dad)

Do one trait at a time, then multiply both.

Blue eyes = gg… what’s the prob. Of gg? ¼ Boy = XY… what’s the prob of XY? ½

¼ x ½ = 1/8 prob of boy w/ blue eyes.

Probability

2. Rule of additiona. Signaled by the word “OR”simplest example: what’s the prob. Of getting a boy OR a girl?

XX x XY prob. of a girl = ½ prob of a boy = ½ ½ + ½ = 1 (100%) Obviously!

Probability

3. Combining the rulesP genotypes: GgTtff x GgTtFfWhat’s the prob. Of a green eyed, short, freckled child?

1. Green eyes:can be GG or Gg

GG = ( ½ x ½) = ¼ Gg = ( ½ x ½ ) + ( ½ x ½ ) = ½ GG( ¼ ) + Gg ( ½ ) = ¾

Probability

3. Combining the rulesP genotypes: GgTtff x GgTtFfWhat’s the prob. Of a green eyed, short, freckled child?

2. Short:can only be tt

tt = ( ½ ) x ( ½ ) = ¼

Probability

3. Combining the rulesP genotypes: GgTtff x GgTtFfWhat’s the prob. Of a green eyed, short, freckled child?

3. Freckled Can be Ff or FF (but looking at the parents, FF will not be a possibility!)

(1) x ( ½ ) = ½

Probability

3. Combining the rulesP genotypes: GgTtff x GgTtFfWhat’s the prob. Of a green eyed, short, freckled child?

1. Green eyes = ¾ 2. Short = ¼ 3. Freckled = ½ ¾ x ¼ x ½ = 3/32

A Closer Look at Meiosis

• Crossing over: Parts of a homologous chromosome cross to the other homologous chromosome.

Recombination

• Case 1: Recombination of genes that are on different chromosomes (Unlinked). YyRr (yellow, round) x yyrr (green, wrinkled)

¼ YyRr (yellow, round) ¼ yyrr (green, wrinkled

¼ Yyrr (yellow, wrinkled) ¼ yyRr(green,round)

½ Parental Pheno. ½ Recombinant Pheno.

Recombination

• Case 1: Recombination of genes that are on different chromosomes (Unlinked).

• 50% recombinant offspring is the EXPECTED value for unlinked genes.

• It is also the MAXIMUM value

Recombination

• Case 2: Recombination of genes that are on the same chromosome (Linked).

• Any number less than 50% means the genes must be on the same chromosome.

• The LOWER the number = the CLOSER the genes are to each other.

Recombination

• Equation for calculating recombination Frequencies & distance the genes are from each other.

Freq. R = # recombinants/ total offspring

Practice Problem

Human Traits & Diseases

1. Pedigree Charts: Diagram used to follow inheritance patterns of genes.

F. Human Disorders

1. Recessive disordersa. Cystic Fibrosis

Symptoms: Excess mucus in lungs, digestive problems, prone to infections.

Most Affected: Northern European ancestry.

F. Human Disorders

1. Recessive disordersa. Cystic Fibrosis

Cause: Deletion of 3 bases = no phenylalanine.

CFTR protein malfunctions = mucus builds up.

Treatment: Nebulizer = Thins mucus in lungs.

F. Human Disorders

1. Recessive disordersa. Cystic Fibrosis

Frequency: 1 in 2,500

Diagnosis: Tested before birth or once symptoms are noticed.

F. Human Disorders

1. Recessive disordersb. Sickle Cell Disease

Frequency: 1 in 400 African Americans

Other races affected but primarily African Americans.

F. Human Disorders

1. Recessive disordersb. Sickle Cell Disease

Causes: Substitution on an incorrect amino acid in hemoglobin protein.

Symptoms: Blood cells are misshapen and clot irregularly.

Treatments: Blood transfusions.

F. Human Disorders

2. Dominant disorders

a. Achondroplasia

Symptoms: Dwarfism, shortened limbs

Most affected: Uncertain; Studies show men being affected more.

F. Human Disorders2. Dominant disorders

a. Achondroplasia

Cause: FGFR3 gene = abnormal cartilage formation.

Treatment: Growth hormones

F. Human Disorders

2. Dominant disorders

a. Achondroplasia

Frequency: Not known in USA; International = 1 in 15,000- 40,000.

Diagnosis: Before birth, after birth.

Sex-Linked Disorders

1. Muscular Dystrophy: Weakening of muscles, loss of coordination. 1/3500 males.

2. Hemophilia: Poor blood clotting, smallest injuries can be fatal.

Chromosome Disorders

1. Nondisjunction: Homologous chromosomes don’t separate during meiosis.

2. Incorrect number of chromosomes.

Chromosome Disorders

3. Down Syndrome

a. 3 copies of chromosome 21

Chromosome Disorders

4. Turner’s Syndromea. One X chromosome = 45,X

b. Sterile, females

5. Klinefelter’s Syndromea. Extra X chromosome = 47, XXY

b. Sterile, males

G. Genetic Testing

1. Testing for allelesa. Test parents

b. Fetal testinga. Amniocentesis

b. CVS

c. Newborn Screening

Influences on Genes

A. Polygenic Traits: Traits that are controlled by more than 1 gene.

1. Eye color

2. Skin color

Influences on Genes

B. The Environment

1. Plant height

2. Human weight