01:447:380 Lecture Notes - Lecture 12: Punnett Square, Phenotypic Trait, Wild Type

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LECTURE  MENDEL’“ LAW“ AND THE PRINCIPLE“ OF HEREDITY
I. VOCAB
a. Phenotypic character--an observable feature, such as flower color
b. Phenotypic trait--a specific form of the character, for example yellow flowers
c. Dominant allele--an allele that will control the phenotype, in spite of the presence of
another allele that would code for a different trait
i. Ex. If A is dominant over a, then the AA genotype will produce the same
phenotype as the Aa genotype
d. Recessive allele--an allele that will only control the phenotype if the individual is
homozygous for that allele
i. Ex. The only way for an individual to exhibit the phenotypic trait that is
associated with the a allele is to be homozygous for the a allele, i.e. have the aa
genotype
e. Hemizygous genotypethe individual only possesses one copy of that chromosome,
and therefore one copy of all the genes that lie on that chromosome
f. True-breeding or pure-breeding--All the idiidual’s a offsprig hae the sae
trait, suggesting that the individual is homozygous for the alleles that code for that
character
g. De novo mutationarises during spermatogenesis or oogenesisthe parents do not
possess it
h. Autosomal mutationsMutations In genes that lie on the numbered chromosomes (1-
22)
i. X-Linked and Y-Linked mutationsMutations in genes that lie on the X or Y
chromosomes
j. Parental (P) generation--The first generation that is mated
k. First filial (F1) generation--The first generation of offspring from the P generation
l. Second filial (F2) generation--The generation after the F1 generation
m. LocusA specific place in the DNA moleculevaries in length
II. Designating Wil-type and variant alleles
a. For every gene, there is one allele that is the most common allele found in the wild--this
is alled the ild-tpe allele, desigated  the letters used for that gee plus a +
sig, or soeties just  the + sig
b. The other alleles of that gene are referred to as variant alleles
c. E: The El gee ifluee the legth of a goat’s ears: ElR = ear length restricted, an allele
that leads to restricted ear length. The wild-type allele = El+
a. A goat that is heterozygous for the two alleles --> El+/ElR or +/ElR
d. PUNNETT SQUARE: shows what genotypes are possible in the ffspring of a mating
i. Put the four possible allele combinations for one parent (in this case, SY, Sy,
sY, sy) across one side of the square, and the possible allele combinations
from the other parent (in this case, the same individual) across the other side
III. Predicting The Probability Of An Outcome
a. AND RULE, i.e. THE MULTIPLICATION RULE FOR INDEPENDENT EVENTS
i. In order to calculate the probability of a set of independent events occurring,
you take the probabilities of the individual events and multiply them together
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1. Ex. The probability of getting 3 heads in a row when flipping a coin:
1/2 probability to get a head on any single flip 1/2 X 1/2 X 1/2 = 1/8
b. OR RULE, I.E. THE ADDITION RULE FOR MUTUALLY EXCLUSIVE EVENTS
i. In order to calculate the probability of a set of mutually exclusive events
occurring, you take the probabilities of the individual events and add them
togetherNote that you first calculate the probability of each individual event
using the AND rule
ii. Ex. The probability of getting three children of the same sex
p(3 girls) = 1/2 X 1/2 X 1/2 = 1/8 (AND rule)
p(3 boys) = 1/2 X 1/2 X 1/2 = 1/8 (AND rule)
p(3 girls OR 3 boys) = 1/8 + 1/8 = 2/8 = 1/4 (OR rule)
c. Rule Of Thumb
i. If there is only one way you can achieve the outcome, use the multiplication
rule; if there are multiple ways you can achieve the outcome, use the addition
rule
ii. Ex. The probability of getting three children of the same sex
p(3 girls) = 1/2 X 1/2 X 1/2 = 1/8 (AND rule)
p(3 boys) = 1/2 X 1/2 X 1/2 = 1/8 (AND rule)
p(3 girls OR 3 boys) = 1/8 + 1/8 = 2/8 = 1/4 (OR rule)
1. There are two ways to get three children of the same sex, so you have
to add p(3 girls) + p(3 boys)
There is only one way to get 3 girls, so multiply to get p(3 girls)
Also multiply to get p(3 boys)
2. For any one child: p(affected) = 1/4 and p(unaffected) = 3/4
Four ways to get one affected and three unaffected children:
Aff, Un, Un, Un = 1/4 X 3/4 X 3/4 X 3/4 = 27/256
Un, Aff, Un, Un = 3/4 X 1/4 X 3/4 X 3/4 = 27/256
Un, Un, Aff, Un = 3/4 X 3/4 X 1/4 X 3/4 = 27/256
Un, Un, Un, Aff = 3/4 X 3/4 X 3/4 X 1/4 = 27/256
Overall probability = 108/256 = 27/64
iii. You can calculate the probability of any combination of affected/unaffected
children using the binomial expansion
1. Recall factorials: (n! = n X n-1 X n- X…)
let p(affected) = 1/4 = p and let the number of affected = s
let p(unaffected) = 3/4 = q and let the number of unaffected = t
Let n = total number of children
2. To calculate the probability of having 2 affected children out of 5 (n =
5, s = 2, t = 3):
p = n! psqt
s!t!
p = 5 X 4 X 3 X 2 X 1 (1/4)2 (3/4)3 = 10 X 1/16 X 27/64 = 270/1024
2 X 1 X 3 X 2 X 1
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Document Summary

Lecture (cid:1005)(cid:1006) mendel" law and the principle of heredity. If a is dominant over a, then the aa genotype will produce the same phenotype as the aa genotype: recessive allele--an allele that will only control the phenotype if the individual is homozygous for that allele, ex. The probability of getting 3 heads in a row when flipping a coin: 1/2 probability to get a head on any single flip 1/2 x 1/2 x 1/2 = 1/8: (cid:862)or(cid:863) rule, i. e. There is only one way to get 3 girls, so multiply to get p(3 girls) Also multiply to get p(3 boys: for any one child: p(affected) = 1/4 and p(unaffected) = 3/4. Four ways to get one affected and three unaffected children: Aff, un, un, un = 1/4 x 3/4 x 3/4 x 3/4 = 27/256. Un, aff, un, un = 3/4 x 1/4 x 3/4 x 3/4 = 27/256.

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