Binomial Expansion Flashcards
Example – If a Heterozygous couple mates – if the couple has 3 kids what is the probability that ONLY 1 child has albanism?
Intuatuvley – you might want to do P(aa) X P(A_) X P(A_) –> This is WRONG
Reason: Now the three events are NOT indepent (Because if have a child with albanism is the first mating even then the next two CANNOT have albanism; if the second kid has albanism then the first and third cannot have albanism – this makes the events dependent on each other
- Wrong because there are multiple ways that the couple can have 3 kids where only one has albansim
MEANS – there are 3 ways that you can have only 1 kid with albanism
1. The first kid has it and the second and thrid don’t
OR
2. The second kid has it and the first and third don’t
OR
3. The third kid has it and the first and second don’t
TO SOLVE:
P(The first kid has it and the second and thrid don’t) + P(The second kid has it and the first and third don’t) + P(The third kid has it and the first and second don’t )
(1/4 X 3/4 X 3/4) + (3/4 X 1/4 X 3/4) + (3/4 X 3/4 X 1/4) = 27/64
OR
Can use Binomial expansion
3 kids – (A + B)^3 –> A^3 + 3a^2b + 3ab^2 + b^3
a = probability of albanism
b = probability of no albanism
THEN – pick the term in bionomal that matches prompt – here it is 1 kid with and 2 without –> ab^2
USE 3ab^2 –> 3 X 1/4 X (3/4)^2 = 27/64
When can you use Binomial expansion
USE if one event determines another (1 kid having a trait effects if the other kids have it – IF uses ONLY = likely use it)
+
NEED 2 categories – can only do if have two phenotypes
Logic for binomial expansion
Need A + B = 1 (ALWAYS TRUE)
Probability of having trait + proabilty of not having trait = 100%
***That pobabilty is the same for each kid
(Probability of having trait + proabilty of not having trait) for kid 1 = (Probability of having trait + proabilty of not having trait) for kid 2
CAN combine terms – (Probability of having trait + proabilty of not having trait) for kid 1AND (Probability of having trait + proabilty of not having trait) for kid 2 –> (Probability of having trait + proabilty of not having trait) for kid 1 X (Probability of having trait + proabilty of not having trait) for kid 2 = (A + B)^2
Biniomial Formula
(A + B) ^n –> N = # of kids
How do you pick the term to use in binomial expansion
Pick the one that matches the prompt – match the defined a and b terms
***KEEP the coefficent in
Example – probability of only 1 kid with albanism and 2 without
a = P(Albanism)
b = P(No albanism)
For three kids – (A + B)^3 –> A^3 + 3a^2b + 3ab^2 + b^3
WANT ab^2 (1 kid with and 2 without) –> 3ab^2
Trick for Binomial Expansion
If can’t expand because don’t remember coefficants –> can use pascals trainge to find the coefficient for each term
***match the exponet in (A + B)^n with line in traiangle
Relationships between alleles is…
CONTEXTUAL – can;t just be domininet allele –> it is an allele that is dominet TO _______
***Important because allele might not be dominent in all cases – might be dominent in one case (dominent to one other allele) but NOT domient to a different allele
EXAMPLE – in image have two alleles that ahve different realtionships depednning on the phenotype
1. In the first row they are co-dominent
2. In the secon one bA is fully domiennet to bS
3. In the third one bA and bS are incompletley domint
Sickle cell vs. normal cell
Normal cell – soft + round
- Blood flows freely in blood vessle
Sickle cell – Hard and cresent shaped
- Blood flow can be blocks
Sickle cell genes
HbA allele – leads to normal RBCs and is dominent to Hbs allele
Hbs allele = leads to sickle cell RBCs
HbA allele in sickle cell
Actually have multiple DNA seq that all lead to same phenotype – all same WT – we put all of the genetic varaition in HbA
- It is the blanket term for all WT alleles (In reality there are multiple WT alleles)
Hemoglobin Structure
Made of 2 polypeptides
1. Alpha Hemoglobin
2. Beta Hemoglobin
Alleles for Beta Hemoglobin
There are many alleles for beta Hemoglobin
ONE allele = HbS –> Gives rise to sickling of the RBCs
DNA seq of Normal RBCs Vs. Sickle cell RBCS (DNA seq for Beta Hemoglobin)
SNP – have one base change – the one base changes causes Gluatamic Acid (normal RBCS) –> Valine (Sickle cell)
- Glutamic acid – shapes + holds heme = normal
- Valine = shift in structure of beta hemolgobin = can’t bind heme = get sickle shape
***Change in AA sequence causes hemoglobin molecules to crystallize when oxygen levels in the blood are low – as a result RBCs sickle and get stuck in Blood Vessels
What is the relationship that descrbes beta A and beta S when looking at B-globin production
ANSWER: Beta A is codimnent to Bs
Reason – the phenotype in questions = Beta globin production (proteins that are produced)
The heteozygous genotype leads – bith versions of beta globine (get the one with heme AND the one without) – because have BOTH dominent and recssive phonetypes NOT some intermediate version = co-dominnet (see both dominent and recssive fully)
- because get both proetins = see both phenotypes = co-dominent
What is the relationship that descrbes beta A and beta S when looking at Red Blood Cell shaoe
ANSWER: bA is dominent to bS
Reason – because the hetrozygous genotype has the same phenotype as the dominent = only dominent = hB is dominent to hS
- because the dominent is normal and the heterozygous is normal = dominent allele
Altitude + sickle cell anemia
Envirnment = can affect phenotype – RBCs at high altitude can change shape – altitude affects the shape
***Envirnment dependent
GENE BY ENVIRNMENT INTERACTIONS
Gene by envirnment interaction
Envirnment influences phenotype (g X e)
***gene whose phenotype depends on relationship between other allles + ARE ALSO influneced by envirnment
Example – sickle cell _ altitude
At high altitude
AA – Normal
AS –> Some sickling
SS –> Sever sicling
VS.
At normal altitude
AA –> Normal
AS –> Normal (because A is completley dominent to S)
SS –> Sickle cell
Example – what is the probability that one of three children has sickle cell – parents are Aa X Aa
Most people want to use 1/4 X 3/4 X 3/4
THIS IS WRONG — because there are three different ways that a couple can have only one kid with and 2 without
1. The first has it and the ither two don’t
2. The second has it
3. The third has it
HERE – the events are NOT independent on each other because if the first one gets it then the other two can’t or if the second one has it then the other two cannot – means that one having it affects the others
HOW TO DO IT:
1. Can combine the probabilities of all of the individual possible events
The first has it and the ither two don’t OR The secon has it and the ither two don’t OR The third has it and the ither two don’t
P(The first has it and the ither two don’t) + P(The second has it and the ither two don’t) + P(The third) has it and the ither two don’t)
(1/4 X 3/4 X 3/4) + (3/4 X 1/4 X 3/4) + (3/4 X 3/4 X 1/4) = 27/64
OR
Can use binomial Expansion
Define
a = pribability of child with
b = probability of child without
(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3
PICK the right term – one with and two withoout – 3ab^2 –> 3(1/4) X (3/4)^2 = 27/64
Pascal’s Triangle + binomial expansion
Can give you the coefifciants for the term
Example – what is the probability of having 5 chidlren – 2 with sickle cell and 3 without (Parents are Aa X Aa)
