Download R r D d

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 Vocabulary
C. 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 things
1.
“Factors” determine what organisms will be
like .
a. Genes – determine traits
1. 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.
2.
Dominant: Will always be seen.
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. Results
1. 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.
Tt
T
TT
Tt
t
T
t
Tt
Tt
tt
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
Homozygous
Dominant
tt = Short
Homozygous
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
TT
Genotype of
parent 2
T
t
tt
t
T
XI. Single Trait Cross
Results: All 4 combinations are the same.
Results vary depending on parents!
T
T
t
Tt
Tt
t
Tt
Tt
XI. Monohybrid Cross
F1 Cross: Cross 2 of the “kids”.
Results: 1:2:1 Genotypic Ratio
3:1 Phenotypic Ratio
T
T
t
t
TT
Tt
Tt
tt
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
TT
Tt
t
Tt
tt
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
Female
Male
X
X
X
XX
XX
50% female
Y
XY
XY
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 X
a.
All X-linked traits will be expressed in males.
Only 1 X chromosome, so either YES or NO!
5. Females = X X
a.
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
RrDd
Genotype of parent 2 = R r D d
Genotype of parent 1 =
1. Figure out the combinations for
each parents’ gametes.
XII. Dihybrid Cross
Genotype of parent 1 =
FIRST
R
RD
r
RrDd
X
D
d
XII. Dihybrid Cross
Genotype of parent 1 =
OUTSIDE
R
r
RrDd
X
Rd
D
d
XII. Dihybrid Cross
Genotype of parent 1 =
INSIDE
R
r
RrDd
X
rD
D
d
XII. Dihybrid Cross
Genotype of parent 1 =
LAST
R
r
RrDd
X
rd
D
d
XII. Dihybrid Cross
Genotype of parent 1 =
RrDd
Gamete combinations =
RD
Rd
rD
rd
XII. Dihybrid Cross
RrDd
Same genotype, same
gamete combinations =
RD
Rd
rD
rd
Genotype of parent 2 =
XII. Dihybrid Cross
2. Make a Punnett square: 16 possibilities
XII. Dihybrid Cross
3. Parents’ possible gametes
RD Rd rD
RD
Rd
rD
rd
rd
XII. Dihybrid Cross
4. Perform Crosses
RD
Rd
rD
rd
R D RRDD
RRDd
RrDD
RrDd
R d RRDd
RRdd
RrDd
Rrdd
r D RrDD
RrDd
rrDD
rrDd
r d RrDd
Rrdd
rrDd
rrdd
XII. Dihybrid Cross
5. Results: 9 = right handed w/dimples
RD
Rd
rD
rd
R D RRDD
RRDd
RrDD
RrDd
R d RRDd
RRdd
RrDd
Rrdd
r D RrDD
RrDd
rrDD
rrDd
r d RrDd
Rrdd
rrDd
rrdd
XII. Dihybrid Cross
5. Results: 3 = Right handed, no dimples
RD
Rd
rD
rd
R D RRDD
RRDd
RrDD
RrDd
R d RRDd
RRdd
RrDd
Rrdd
r D RrDD
RrDd
rrDD
rrDd
r d RrDd
Rrdd
rrDd
rrdd
XII. Dihybrid Cross
5. Results: 3 = Left handed w/dimples
RD
Rd
rD
rd
R D RRDD
RRDd
RrDD
RrDd
R d RRDd
RRdd
RrDd
Rrdd
r D RrDD
RrDd
rrDD
rrDd
r d RrDd
Rrdd
rrDd
rrdd
XII. Dihybrid Cross
5. Results: 1 = Left handed, no dimples
RD
Rd
rD
rd
R D RRDD
RRDd
RrDD
RrDd
R d RRDd
RRdd
RrDd
Rrdd
r D RrDD
RrDd
rrDD
rrDd
r d RrDd
Rrdd
rrDd
rrdd
XII. Dihybrid Cross
5. Results:
The probability of having a child with dimples
and being left-handed?
3/16 =
rrDD
rrDd
rrdD
Out 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 multiplication
a. 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 = gg
GgXX (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 addition
a. 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 rules
P genotypes: GgTtff x GgTtFf
What’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 rules
P genotypes: GgTtff x GgTtFf
What’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 rules
P genotypes: GgTtff x GgTtFf
What’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 rules
P genotypes: GgTtff x GgTtFf
What’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
Statistical Analysis
• Data can be misleading
• How certain are you that your data are
statistically significant?
Statistical Analysis
• Null hypothesis = There is no statistically
significant difference between observed
data and expected data.
• Alternate hypothesis = There is a
statistically significant difference between
observed data and expected data.
Statistical Analysis
•
•
•
•
Methods: Chi square analysis
X2 = Σ [(o – e)2 / e)]
Σ = sum
o = observed # individuals
e = expected # individuals
Statistical Analysis
• Example Problem
Round is dominant to wrinkled in pea seeds.
A cross produced 722 round seeds and 278
wrinkled seeds.
Is this data significant enough to verify the
expected inheritance pattern?
Statistical Analysis
phenotype # observed # expected (o – e)
(o)
(e)
green
albino
(o – e) 2
(o – e) 2 / e
Statistical Analysis
Step 1: Calculate Chi Square
• X2 = Σ [(o – e) 2 / e] = ____________
Statistical Analysis
Step 2: Determine degrees of freedom (df)
- The # phenotypes minus 1
- df = ___________
Statistical Analysis
Step 3: Find critical value at .05 p value
Probability
(p)
.05
Degrees of Freedom (df)
1
2
3
4
5
3.84
5.99
7.82
9.49
11.1
Statistical Analysis
Step 4: Use critical values to analyze results
- IF Chi square value is > or equal to critical
value, the null hypothesis is REJECTED.
- Meaning… the data are not statistically
significant.
- IF Chi square value is < the critical value,
the null hypothesis is ACCEPTED.
-Meaning… The data are statistically
significant.
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.
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
Human Traits & Diseases
1. Pedigree Charts: Diagram used to follow
inheritance patterns of genes.
F. Human Disorders
1. Recessive disorders
a. Cystic Fibrosis
Symptoms: Excess mucus in lungs, digestive
problems, prone to infections.
Most Affected: Northern European ancestry.
F. Human Disorders
1. Recessive disorders
a. 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 disorders
a. Cystic Fibrosis
Frequency: 1 in 2,500
Diagnosis: Tested before birth or once
symptoms are noticed.
F. Human Disorders
1. Recessive disorders
b. Sickle Cell Disease
Frequency: 1 in 400 African Americans
Other races affected but primarily African
Americans.
F. Human Disorders
1. Recessive disorders
b. 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 Disorders
2. 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 Syndrome
a. One X chromosome = 45,X
b. Sterile, females
5. Klinefelter’s Syndrome
a. Extra X chromosome = 47, XXY
b. Sterile, males
G. Genetic Testing
1. Testing for alleles
a. Test parents
b. Fetal testing
a.
b.
c.
Amniocentesis
CVS
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