BIOL 1090 Lecture Notes - Lecture 6: Binomial Distribution, Zygosity
21 Aug 2016
School
Department
Course
Professor
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Ex: if dominant phenotype --> dunno if homozygous dominant or heterozygous!
Breed unknown genotype with homozygous recessive!
If any offspring is recessive --> we know parent was heterozygous!
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Can find out using test cross!
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We can't always tell genotype by just looking!
So we use pedigrees!
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BUT --> we can't do that to people :P
Square = male
Circle = female
Diamond = unknown sex
Coloured = display trait
Roman numerals = generation
Numbers beneath shapes = individuals within generation
Sometimes use shorthand numbers inside shapes (number of children of indicated sex
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Many notations used!
Appears every generation! (very unlikely for spontaneous appearance!)
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Every affected individual has at least one affected parent
1.
Dominant Trait Inheritance
"Skips" a generation! (every other!)
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Trait can suddenly appear in a pedigree!
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In absence of evidence to the contrary, assume that individuals marrying into family
do not carry recessive allele!
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Standard assumption:
We know of 4000+ recessive human traits
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2.
Recessive Trait Inheritance
Many types of inheritance!
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Pedigrees show relationships between members of a family
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Mendel's laws still apply!! But just not as predictive with smaller family sizes
Because human families are relatively small, the phenotypic ratios often deviate significantly from Mendelian
expectations!
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Example 1:
There are 5 outcomes! (some have multiple ways to get to it)
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Consider a couple, both heterozygous for recessive disease. What's the chance of 1 of their children having it,
if they have 4 children?
Outcome
#
Overall
outcome
Sample points
Calculation
Probability
1
4U, 0A
{UUUU}
1 x (3/4)^4
81/256
2
3U, 1A
{UUUA, UUAU, UAUU, AUUU}
4 x (3/4)^3 x (1/4)
108/256
3
2U, 2A
{UUAA, UAAU, AAUU, AUAU, UAUA, AUUA}
6 x (3/4)^2 x (1/4)^2
54/256
4
1U, 3A
{UAAA, AAAU, AUAA, AAUA}
4 x (3/4) x (1/4)^3
12/256
5
0U, 4A
{AAAA}
1 x (1/4)^4
1/256
P(A) = P(cc) = 1/2 x 1/2 = 1/4
P(U) = P(CC) + P(Cc) = 1/4 + 1/2 = 3/4
Treat each birth as individual:
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= (3/4)^3 x (1/4)
= 108/256
= 42% chance! (most likely outcome)
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P(3U, 1A) = P(U) x P(U) x P(U) x P(A)
Mathematically, for a total number of nprogeny, we can calculate the binomial probability that x progeny
will fall in one (P) class and y progeny will fall in the other (Q) class as:
BIOL*1090-01
Lecture 6 - Pedigrees & Human Genetics
January 27, 2016
8:30 AM
BIOL 1090 Page 1
Document Summary
We can"t always tell genotype by just looking! Ex: if dominant phenotype --> dunno if homozygous dominant or heterozygous! If any offspring is recessive --> we know parent was heterozygous! But --> we can"t do that to people :p. Pedigrees show relationships between members of a family. Numbers beneath shapes = individuals within generation. Sometimes use shorthand numbers inside shapes (number of children of indicated sex. Many types of inheritance: dominant trait inheritance. Appears every generation! (very unlikely for spontaneous appearance!) Every affected individual has at least one affected parent: recessive trait inheritance. In absence of evidence to the contrary, assume that individuals marrying into family do not carry recessive allele! Because human families are relatively small, the phenotypic ratios often deviate significantly from mendelian expectations! But just not as predictive with smaller family sizes. Consider a couple, both heterozygous for recessive disease. There are 5 outcomes! (some have multiple ways to get to it)
