Transcript Basic Mendelian Crosses

Basic Mendelian Crosses
Purebreeding Parents
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Basic Mendelian Crosses
Testcross
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Basic Mendelian Crosses
F1 x F1 Cross
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Forked Line (Branch) Diagrams
Gametes possible from AaBbccDd individual
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Forked-Line (Branch) Diagrams
Phenotypes possible
from cross:
AaBbCcDd x AaBbCcDd
p(A- B- C- D-) = 81/256
P(A- bb C- dd) = 9/256
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Forked-Line (Branch) Diagrams
Genotypes possible from cross Aa Bb CC x Aa bb cc
1/2 Bb
1/1 Cc
1/2 bb
1/1 Cc
1/2 Bb
1/1 Cc
1/2 bb
1/1 Cc
1/2 Bb
1/1 Cc
1/2 bb
1/1 Cc
1/4 AA
p (AA bb Cc)?
1/2 Aa
p (aa bb Cc)?
1/4 aa
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Mathematical Method
Answers specific question, not all inclusive
Individual with genotype Aa Bb cc Dd
What is the probability of gamete with abcd ?
p (a) = 1/2
p (b) = 1/2
p (c) = 1
p (d) = 1/2
p (a) and (b) and (c) and (d) = 1/2 * 1/2 * 1 * 1/2 = 1/8
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Mathematical Method
Cross between Aa Bb Cc x Aa Bb Cc
What is the probability of offspring with genotype:
Aa BB cc?
p (Aa) = 1/2
p (BB) = 1/4
p (cc) = 1/4
p (Aa) and (BB) and (cc) = 1/2 * 1/4 * 1/4 = 1/32
AA BB Cc?
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Mathematical Method
Cross between Aa Bb Cc Dd EE x aa Bb CC dd Ee
p (aa bb CC dd Ee)?
p (Aa Bb Cc Dd Ee)?
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Mathematical Method
When ratios for all genes are consistent:
Ex. Aa Bb Cc x Aa Bb Cc
# Phenotypes = (2)n
2 = # phenotypes for each gene
n = # monohybrid crosses (genes)
Ex. (2)3 = 8
# Genotypes = (3)n
3 = # genotypes for each gene
n = # monohybrid crosses (genes)
Ex. (3)3 = 27
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Calculating Number of Genotypes Possible
# genotypes possible = 3 n
3 = Genotypes possible for each gene - AA, Aa, aa
n = # heterozygous gene pairs
Calculating Number of Gametes Possible
# gametes possible = 2 n
2 = diploid with 2 copies of each gene/chromosome
n = # heterozygous gene pairs
Ex. AaBb :
2 2 = 4 possible (AB, Ab, aB, ab)
Ex. AaBbCc :
2 3 = 8 possible (ABC, ABc, Abc, etc.)
Human chromosomes :
2 23 = > 8 x 10 6
Binomial Expansion: Uses
Predict comprehensive phenotypic ratios ( 9:3:3:1, etc.)
Determine probability of particular categories
Only applicable if:
Ratio for every trait (gene) is the same (ex. AaBb x AaBb - 3:1)
Not applicable if:
Ratios vary (ex. AaBb x Aabb - 3:1 for A-:aa; 1:1 for B-:bb)
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Binomial Expansion: Generating Comprehensive Ratios
Probabilities: p = dominant phenotype, q = recessive phenotype
N = # genes
X = # of dominant
(N-X) = # of recessives
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Binomial Expansion: Generating Comprehensive Ratios
Example based on phenotypes from AaBb x AaBb cross
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A-B-: A-bb: aaB-: aabb
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Binomial Expansion: Generating Comprehensive Ratios
Example for AaBbCcDdEe x AaBbCcDdEe cross
(p+q)5 = p5 + 5 p4q + 10 p3q2 + 10 p2q3 + 5 pq4 + q5
5!
4! 1!
=
5*4*3*2*1 =
(4*3*2*1) * 1
5!
3! 2!
=
5* 4 * 3 * 2 * 1 =
(3 *2*1) * (2*1)
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Binomial Expansion: Ways to Determine Coefficients
Pascal’s Pyramid
p
p6 +
q
p5q + p4q2 + p3q3 + p2q4 + pq5 + q6
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Binomial Expansion: Ways to Determine Coefficients
Direct method (shortcut):
Example: (p+q)4 =
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Binomial Expansion: Determining Phenotypic Ratios
Probability: p(A-B-C-) = 27/(43) = 27/64
p (A-B-cc) =
p(two dominant and one recessive) =
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Binomial Expansion: Determining Phenotypic Ratios
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Binomial Expansion: Determining Probability
p = probability of one event (ex. girl)
q = probability of alternative event (ex. boy)
Probability that in N trials, you will get X girls and (N-X) boys
=
N!
(pX) (q)(N-X)
X! (N-X)!
Note: Only use coefficient
when order is not considered
Example: probability of 2 girls and 4 boys in a family of 6?
6! (1/2)2 (1/2) 4 =
2! 4!
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Examples of Applying Binomial
Two brown-eyed parents mate and have a blue-eyed child.
What are the parents genotypes?
Which allele for color is dominant?
If two individuals with the same genotype had four children,
what is the probability of them having all blue eyes?
What is the probability of them having two brown and two blue?
What is the probability of the first two being brown and the rest blue?
In how many different orders could this occur?
What is the probability of their fifth child having blue eyes?
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Sample Problem
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Answer to Sample Problem (Part 1)
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Answer to Sample Problem (Part 2)
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Pedigree Analysis
Symbols Used
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Pedigree Analysis
Sample Pedigree
For rare conditions:
assume those outside
family are homozygous
or hemizygous normal;
Genetic counselor:
p (carrier in population)
ex. CF carrier = 0.05
Arrow indicates propositus
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Pedigree for Autosomal Recessive Trait
Usually loss-of-function
Albinism: absence of pigment in skin, eyes, hair
Cystic fibrosis: thick mucus that blocks lungs, glands
Sickle cell anemia: abnormal hemoglobin, blockage
Xeroderma pigmentosum: No nucleotide excision repair
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Things to Look for with Autosomal Recessive Traits
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Pedigree for Autosomal Dominant Trait
Insufficient product, interference with normal, or gain-of-function
Achondroplasia: dwarfism, defect in long bone growth
Brachydactyly: shortened fingers
Hypercholesterolemia: high cholesterol, heart disease
Huntington disease: nervous system degeneration
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Things to Look for with Autosomal Dominant Traits
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Are these autosomal traits dominant or recessive?
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Are these autosomal traits dominant or recessive?
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Are these autosomal traits dominant or recessive?
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Are these autosomal traits dominant or recessive?
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Are these autosomal traits dominant or recessive?
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Pedigrees for X- linked Traits
Recessive
Color blindness: insensitivity to red or green light
Hemophilia: defective clotting, A or B type
Muscular dystrophy: Duschenne type, muscle wasting
Dominant
Hypophosphatemia: rickets (bowlegged)
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What to look for with X- linked Traits
Dominant
XAXA or XAXa
Recessive
XaXa
Hemizygous
XAY
XaY
Recessive
Dominant
Affected daughter
has affected father
and carrier mother
Affected father
passes it on to
all daughters
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What to look for with X- linked Recessive Traits
Mother can carry and pass on the recessive allele
More males express these traits
Mother to son inheritance
Affected daughters must have affected fathers
Criss-cross pattern of inheritance
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Sample X- linked Recessive Pedigree
X = normal, X = abnormal
Males: 1/2 normal, 1/2 affected
Females: 1/2 normal, 1/2 carriers
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Sample X- linked Recessive Pedigree
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What to look for with X- linked Dominant Traits
Affected mother will pass it on to 50% both daughters and sons
Affected father will only pass it on to all daughters
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Sample X- linked Dominant Pedigree
X = normal, X = abnormal
Overall: 1/2 affected, 1/2 normal
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Y- linked (Holandric) Pedigrees
Traits are only seen in males and passed on from father to son.
X = normal, Y = abnormal
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Sample Pedigree Problem
Adherent Earlobes (recessive autosomal)
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Genotypes: II 3 =
p II 1 is Aa =
I 1 and 2 =
p III 2 is Aa =
p III 2 and III 4 would have aa child =
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Sample Pedigree Problem
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Sample Pedigree Problem
Huntington Disease (same family, later date)
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Sample Pedigree Problem
Autosomal trait (rare recessive)
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Sample Pedigree Problem from Lab
I
II
III
Probability of carrier?
1-4
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8
9
15
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If 16 married 17, what is the probability of an affected child?
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