unit 4: cell division & heredity
DESCRIPTION
Unit 4: Cell Division & Heredity. Part 3 Ch. 14 & Ch. 15 Mendelian Genetics & Chromosomal Basis of Inheritance. I. Genetics As A Science. Genetics = Study of heredity Heredity = How information gets transferred to us from our parents. II. Genetics Vocabulary. - PowerPoint PPT PresentationTRANSCRIPT
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Unit 4: Cell Division & Heredity
Part 3
Ch. 14 & Ch. 15
Mendelian Genetics & Chromosomal Basis of Inheritance
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I. Genetics As A Science
A. Genetics = Study of heredity
1. Heredity = How information gets transferred to us from our parents.
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II. Genetics Vocabulary
A. Trait: Characteristic that is different among individuals.
Eye color, height, skin color
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II. Genetics Vocabulary
B. Cross: When 2 individuals mate.
2 trees, 2 dogs, etc.
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II. Genetics VocabularyC. Hybrid: Result of crossing individuals with
different traits.
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III. Discovery of Genetics
A. Gregor Mendel: Used pea plants to explore heredity.
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III. Discovery of Genetics
B. Mendel concluded 2 things1. “Factors” determine what organisms will be
like .
a. Genes – determine traits1. Height
2. Hair color
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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
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III. Discovery of Genetics
B. Mendel concluded 2 things…
2. Some alleles are “stronger” than others
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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.
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IV. Mendel’s Work
A. Crossed a tall plant with a short plant.
1. P Generation = Original pair of plants.
(Parents)
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IV. Mendel’s Work
B. Results1. All tall plants
F1 = First set of offspring.
(children)
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IV. Mendel’s Work
C. Curious about why none were short.
1. Crossed 2 plants from the F1. generation.
Yes, mated 2 siblings!
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IV. Mendel’s Work
D. Results
1. 75% Tall
25% Short
F2 = offspring of the F1 generation.
(grandchildren)
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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”?
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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
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VIII. More About Alleles
A. Allele combinations
1. Have 2 alleles for each gene
Height – Tall or short
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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)
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VIII. More About Alleles
a. If 2 of the SAME allele: HOMOZYGOUS
TT = Tall tt = Short
Homozygous Homozygous
Dominant Recessive
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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
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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
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X. Punnett Squares
A. Diagrams for predicting results of any cross.
Genotype of parent 1
Genotypes of
offspring
Genotype of
parent 2
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XI. Single Trait Cross
A. A cross for 1 trait.
1. Hair color ONLY
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XI. Single Trait Cross
Genotype of parent 1
T T
Genotype of
parent 2
t t
T T
t
t
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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
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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
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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
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XII. Dominant/Recessive??A. Simple rules of
dominance/recessive don’t always hold true.
1. Incomplete dominance:
Both alleles present = Intermediate
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XII. Dominant/Recessive??
2. Codominance:
Both alleles present = Both appear
Brown & White
both dominant =
both colors
appear
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XII. Dominant/Recessive??3. Multiple alleles:
More than 2 alleles for 1
gene
Blood Groups
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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
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Sex Chromosomes
2. 44 Autosomes & 2 Sex chromosomes
a. Male: 46,XY b. Female: 46,XX
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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
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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
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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.
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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
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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
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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
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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)
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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.
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XII. Dihybrid Cross
Genotype of parent 1 = R r D dFIRST
R r X D d
R D
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XII. Dihybrid Cross
Genotype of parent 1 = R r D dOUTSIDER r X D d
R d
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XII. Dihybrid Cross
Genotype of parent 1 = R r D dINSIDER r X D d
r D
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XII. Dihybrid Cross
Genotype of parent 1 = R r D dLASTR r X D d
r d
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XII. Dihybrid Cross
Genotype of parent 1 = R r D d
Gamete combinations =
R D R d r D r d
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XII. Dihybrid Cross
Genotype of parent 2 = R r D d
Same genotype, same
gamete combinations =
R D R d r D r d
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XII. Dihybrid Cross2. Make a Punnett square: 16 possibilities
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XII. Dihybrid Cross
3. Parents’ possible gametes
R D R d r D r d
R D
R d
r D
r d
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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
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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
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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
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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
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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
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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.
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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?
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XII. Dihybrid Cross
6. Independent Assortment
d. On different chromosomes!
* Different genes can assort without affecting each other if they are on different chromosomes.
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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…
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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?
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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.
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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!
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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
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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
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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
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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
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Statistical Analysis
• Data can be misleading
• How certain are you that your data are statistically significant?
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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.
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Statistical Analysis
• Methods: Chi square analysis
• X2 = Σ [(o – e)2 / e)]
• Σ = sum o = observed # individuals
• e = expected # individuals
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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?
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Statistical Analysis
phenotype # observed (o)
# expected (e)
(o – e) (o – e) 2 (o – e) 2 / e
green
albino
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Statistical Analysis
Step 1: Calculate Chi Square
• X2 = Σ [(o – e) 2 / e] = ____________
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Statistical Analysis
Step 2: Determine degrees of freedom (df)
- The # phenotypes minus 1
- df = ___________
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Statistical Analysis
Step 3: Find critical value at .05 p value
Probability (p)
Degrees of Freedom (df)
1 2 3 4 5
.05 3.84 5.99 7.82 9.49 11.1
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Statistical AnalysisStep 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.
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A Closer Look at Meiosis
• Crossing over: Parts of a homologous chromosome cross to the other homologous chromosome.
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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.
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Recombination
• Case 1: Recombination of genes that are on different chromosomes (Unlinked).
• 50% recombinant offspring is the EXPECTED value for unlinked genes.
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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.
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Recombination
• Equation for calculating recombination Frequencies & distance the genes are from each other.
Freq. R = # recombinants/ total offspring
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Human Traits & Diseases
1. Pedigree Charts: Diagram used to follow inheritance patterns of genes.
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F. Human Disorders
1. Recessive disordersa. Cystic Fibrosis
Symptoms: Excess mucus in lungs, digestive problems, prone to infections.
Most Affected: Northern European ancestry.
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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.
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F. Human Disorders
1. Recessive disordersa. Cystic Fibrosis
Frequency: 1 in 2,500
Diagnosis: Tested before birth or once symptoms are noticed.
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F. Human Disorders
1. Recessive disordersb. Sickle Cell Disease
Frequency: 1 in 400 African Americans
Other races affected but primarily African Americans.
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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.
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F. Human Disorders
2. Dominant disorders
a. Achondroplasia
Symptoms: Dwarfism, shortened limbs
Most affected: Uncertain; Studies show men being affected more.
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F. Human Disorders2. Dominant disorders
a. Achondroplasia
Cause: FGFR3 gene = abnormal cartilage formation.
Treatment: Growth hormones
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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.
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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.
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Chromosome Disorders
1. Nondisjunction: Homologous chromosomes don’t separate during meiosis.
2. Incorrect number of chromosomes.
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Chromosome Disorders
3. Down Syndrome
a. 3 copies of chromosome 21
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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
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G. Genetic Testing
1. Testing for allelesa. Test parents
b. Fetal testinga. Amniocentesis
b. CVS
c. Newborn Screening
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Influences on Genes
A. Polygenic Traits: Traits that are controlled by more than 1 gene.
1. Eye color
2. Skin color
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Influences on Genes
B. The Environment
1. Plant height
2. Human weight