Advanced Pedigree Analysis

Question 1

Describe features that will indicate Autosomal Dominant Inheritance in a pedigree

Answer 1

Autosomal Dominant Inheritance:
- Affected individuals are seen in every generation.
-Half of the children of a heterozygous parent, will be affected

(you may also notice affected daughter, who has unaffected daughter)

Question 2

What does a pedigree for autosomal recessive look like?

Answer 2

Autosomal Recessive Inheritance
-SKIPS generations
-can be consaguinous marriages (among second cousins)
-An affected child will have parents that are BOTH Carriers
(affected child equally likely to be boy or girl)

Question 3

What does a pedigree for sex-linked dominant Inheritance look like?

Answer 3

Sex-linked Dominant Inheritance
-All generations are affected
-if fathers are affected, ALL daughters will be affected (NOT sons, since they get Y from dad)
-if mother is affected and heterozygous she can, pass on mutated gene to son and daughter (or normal copy to son or daughter; hence mix)

Question 4

Describe the features of a pedigree that is sex-linked recessive

Answer 4

Sex-linked Recessive Inheritance
-Skips generations
-Males with trait are More COMMON than females
A woman who is a carrier, half of sons will get trait (mutated copy)

Question 5

What kind of features are seen in pedigree with Mitochondrial inheritance ?

Answer 5

Mitochondrial inheritance
-if mother is affected, all children will be affected.
(variation in the number of defective mitochondria vs normal mitochondria)

Question 6

What are examples of genes with multiple alleles?

Answer 6

Genes with multiple alleles
-The ABO blood group
-The MHC (major histocompatibility) genes

Question 7

Explain the different blood types and antigens that are involved

Answer 7

ABO blood types are based on red blood cell antigens
-red blood cells produce a protein that is modified by the ABO gene to have different sugar groups hanging off the end.
Blood type A; will develop antibodies that attack B; A antigen (in RBC)
-Blood type B: will have antibodies that attack A ; B antigen (in RBC)
-Blood type AB: will have NO antibodies in plasma. They have A and B antigens. **Hence they can receive blood from A, B, and O (any type)
Blood type O: have Antibodies that attack A, B and AB blood. They have no antigens in red blood cell. They can only receive from O. No modifications of protein.

(H; h gene determines whether you make protein and if it can be modified)

(AB half proteins modified one way; half modified another way)

Question 8

What groups are included in structures of A, B and O antigens?

Answer 8

-The A-type antigen ends in N-acetyl galactosamine
-The B-type antigen ends in galactose
-The O-type has NEITHER group at the end

Question 9

Which common alleles are the ABO blood types controlled by? what are the phenotypes for these alleles?

Answer 9

ABO blood types are controlled by three common alleles
-The ABO gene has three alleles; I^A, I^B and I
-The I^A and I^B are co-dominant, since each make a distinct antigen
-The i allele is recessive and makes NO Antigen
Thus, there are four possible phenotypes:
- I^A I^A individuals are type A
-I^B I^B individuals are type B
-I^A I^B individuals are type AB
- ii individuals are type O

Question 10

What are MHC genes? What is their role?

Answer 10

Major Histocompatibility genes (MHC)
- seen in human cells that have special receptors that present antigens to cell of immune system.
MHC proteins are on surface of cells. (function: They grab a protein fragment and hold it so when killer cell, B cell or Helper T cell comes by, the cell will have receptors that recognize antigen determine if it causes infection or if harmless)
MHC genes play an important role in the immune response

Question 11

Where are MHC genes seen? What is their purpose?

Answer 11

Chromosome 6 contains a cluster of MHC genes
- These genes encode receptors that present antigens to the immune system
-differences in these receptors lead to transplant rejections
-HLA MHC complex: set of genes
(the alleles you have on genes, determine what antigen immune system recognizes)
(hence the chance of transplant mismatches are common, since body will recognize new protein in receptor as foreign and attack transplant)

Question 12

How many alleles do HLA genes have?

Answer 12

HLA genes each have large numbers of different alleles
ex: HLA-A have only 68 common alleles
Hence, for HLA-A gene, most people will not have identical pairs of alleles for HLA-A gene.
(So chance of finding match for organ transplant is difficult )
A lot of genetic variation for MHC genes

Question 13

What are the types of errors that can be seen on a pedigree?

Answer 13

Types of errors:
-Some information in the pedigree might be MISSING
-Unknown relationships
-Unknown phenotypes
-The information in the pedigree might be INCORRECT
-Incorrect familial relationships
-Incorrect phenotypes

Question 14

Explain the ethical standards that is required in genetic counseling

Answer 14

Genetic counseling relies on adherence to ethical standards
-Discuss possible situations with your clients before they arise
-Do not make decisions for your clients; give them information and allow them to choose
-Be honest and clear

Question 15

What is conditional probability ?

Answer 15

Conditional probability; measuring the likelihood of an event or outcome occurring based on the occurrence of a previous event or outcome

Question 16

What is the Monty Hall problem? what does it discuss

Answer 16

The Monty Hall problem
national debate about probability and statistics.
-On Let’s Make a deal contestants chose one of three doors, hoping for the one with a car
-After the choice, Monty hall would open one of the remaining doors, revealing a goat
-He then offered the chance to switch choices (after initially choosing door)

Question 17

Explain if you should swtich doors? after choosing one Initially.

Answer 17

Should you switch?
After initially choosing door, and reveals goat (not what you want).
-initally, 1/3 chance of opening door with prize (car)
Is it better to switch or not?
you should SWITCH door during second chance because the two doors are no longer equally likely (you have been given extra information)
-if you pick the correct door: 1/3 chance
-he will open one of two remaining doors, you get wrong
if you did not open right door initially, and pick goat, the man will have to reveal other goat and switching wins.
(you choose bad door 2 or bad door 3; so you switch win, other door becomes 2/3 (odds of remaining door, went up by info being revealed)

Question 18

What is the math equation used for conditional probability problems?

Answer 18

Conditional probability
-When we get new information, we must revise probabilities
-The math used for these type of problems is called BAYESIAN ANALYSIS
Equation:
P (B given A)= P(A if B)*P(B)/P(A)

(like changing piece of pie)

Question 19

what are examples of how Bayesian Analysis is sued in Genetics?

Answer 19

Bayesian Analysis in Genetics:
EX: Dominant mutations that recessive lethal
-The dominant mutation that causes polycystic kidney disease is also recessive lethal
-If told that two people with polycystic kidney disease conceive and carry to term a health baby, what is the chance that baby is genetically normal?
P (child is normal)= 1/3
-exclude the PP out of 4 genotypes (that is dominant allele; but recessive ) and will cause death. So only 1/3 will be normal
(pp would be recessive, hence normal)

equation : P (B given A)= P (B if A) X P(B)/P(A)
P= P (normal given that it was born)* P(normal)/ P(born)
P= P (1) (1/4)/(3/4)

Question 20

What is another example that uses Bayesian analysis ?

Answer 20

Ex; Using children to evaluate the probability of the mother’s genotype

Question 21

What kind of variations can be seen in phenotype?

Answer 21

Variations in phenotype:
-Penetrance and expressivity

Question 22

What occurs in penetrance? Provide an example

Answer 22

Penetrance: what fraction of people who have genotype show the trait
-some genotypes do NOT always cause a specific phenotype
-for example, not all women with a dominant BRCA1 mutation develop cancer
-We say “BRCA1 mutations are partially penetrant””
Or more precisely “BRCA1 mutations are 60% penetrant”
-If a gene is partially penetrant, we can’t always tell the genotype be looking at someone

Question 23

Explain what occurs in expressivity in terms of genetics

Answer 23

Expressivity
-some genotypes cause a variety of stronger or weaker phenotypes
(although 2 individuals have same genotype)
We call these variations in differences in Expressivity
-Sometimes, differences in expressivity can make it unclear if you are studying one or two genes
expressivity is very common

Question 24

What is an example of penetrance and expressivity?

Answer 24

BB (purple color) X bb (white color) can lead to Bb heterozygotes (heterozygous should all be genetically identical)
-however the expressivity presents wide range of differences in phenotype
-Variable penetrance - some show dominant trait (purple), others to
variable expressivity ; same genetic trait, but some show trait more severely, some show less severely.
-Variable penetrance and expressivity; get complicated pattern, although they are all genetically identical

Question 25

What is observed in genetics of human populations?

Answer 25

The STABILITY OF allele frequencies

Question 26

Explain how allele frequencies can be measured. What equations are involved?

Answer 26

Some allele frequencies can be measured DIRECTLY
example; frequencies of blood type (MM, MN, NN) in different populations.
you can notice in data that as migration occurred, there were small variations in frequencies of different subpopulation that moved to different places; give rise to large differences in frequencies of alleles in population of new places.
**once we know genotypes, you can figure out allele frequencies using Hardy Weinberg equation

Question 27

What are the following things that must be true to use Hardy Weinberg equation?

Answer 27

Hardy Weinberg Equation:
Used if the following things are true:
-The population is Large
-Mating is random
-There is NO selective benefit to either allele
-There are NO new mutations
-The gene is not on the X
if these are true, then we can predict genotype frequencies:
f(MM) = f (M)^2
F (MN)= 2f (M)f(N)
F (NN)= f (N)^2

the frequency of M allele is MM + 1/2 of heterozygotes
(can think of M as (p) and N (as q) )
(

Question 28

How else can you write Hardy Weinberg equation?

Answer 28

Hardy Weinberg Equation
-Since the frequency of all allele adds up to 1, f(M) + f(N)= 1
-Since the frequency of all genotypes adds to 1, f(MM) +f(MN)+ f(NN)=1
-You can substitute in Hardy Weinberg values:
f(M)^2 + 2f (M)f(N)+f(N)^2
- OR as you have probably seen it written before:
p^2+2pq+q^2 = 1

Question 29

What can be predicted from allele frequencies ?

Answer 29

GENOTYPE Frequencies can be predicted from allele frequencies
- we can measure frequency of each genotype of each blood type
-Within populations, the expected values from Hardy Weinberg equation are almost identical to observed values
(hence why HW rules must be applied)

Question 30

How do you use the Hardy Weinburg equation?

Answer 30

Using the Hardy-Weinberg Equation
-We generally work backwards
- if the distribution of genotypes follows the H-W equation, then we assume all of the conditions were met:
-Large population, random mating, NO selection
-If the distribution of genotypes is Unexpected, then one of these conditions is NOT correct

Question 31

Explain how Fitness of alleles are measured and how its affects Human Health

Answer 31

Measuring Fitness
-Each genotype has a fitness; the chance of surviving and reproducing
(based on direct observation)
-A fitness of 1 means that genotype has the BEST chance of surviving and reproducing
-A fitness of 0 means that genotype will NEVER survive and reproduce
-The fitness of AA= WAA
(W is fitness)
-The selective coefficient of AA= 1- WAA
(1- WAA; fitness or how much selection you get against allele)

Question 32

Explain what occurs in the fitness of Dominant lethal

Answer 32

Ex; Dominant Lethal
-After a shift to a new environment, A is dominant lethal (disease will kill the person affected)
Hence everyone with A is susceptible to disease. Meanwhile everyone with a is not harmed
if originally, frequency of big A allele = 0.7 and frequency of little a allele= 0.3. Then in new environment, fitness: WAA= 0 , WAa = 0 and Waa= 1
we find out that at birth that:
f(AA) = (0.7)^2 = 0.49
f(Aa) = 2 (0.7)(0.3)= 0.42
f(aa)= (0.3)^2=0.09
Then new virus has come and changed environment; so multiply fitness value by frequency at birth to get frequency at reproduction
f(AA)= 0.49 x 0 (fitness)= 0
f(Aa)= 0.42 x 0 (fitness)= 0
f(aa)= 0.09 x 1 (fitness) = 0.09
hence those with AA and Aa all die before reproduction
The frequency at reproduction: 0+0+0.09 will sum to 0.09.
then to get Normalized value ; frequency at reproduction for each genotype divided by 0.09.
**Hence new f(a)= 1 after only a single generation of selection

(selecting dominant and harmful and common is extremely powerful)
-the dominant mutations that are harmful are rare, and have frequent mutations that lead to dominant mutations.

Question 33

Explain what occurs in frequencies with Common Recessive Lethal

Answer 33

Common Recessive Lethal
-After a shift to a new environment, b is recessive lethal
if f(B)= 0.1, f(b)-0.9, and W BB=1, W Bb=1 and W bb=0

Genotype F(birth) Fitness F(Repro) Norm
f (BB) (0.1)2=0.01 x 1 0.01 0.05
f(Bb) 2(0.1)(0.9)= 0.18 x 1 0.18 0.95
f (bb) (0.9)^2= 0.81 x 0 0.00 0.00

Total——————————>0.19
The new f(b)= 0.5 (0.95) =0.475
1/2 value of heterozygous x normalized value of Bb
-A huge decrease after a single generation of selection
Normalized values: divide each frequency value at reproduction divided by the summation of freq at reproduction (ex: Norm for F(BB)= 0.01/0.19= 0.05

Question 34

Explain what occurs in frequencies of Rare recessive Lethal

Answer 34

Rare recessive Lethal
Cystic fibrosis is the most COMMON genetic disorder among caucasians
f(C)= 0.98, f (c)=0.02, Assume WCC=1, WCc= 1 and Wcc = 0
(That is a worst case scenario, give modern treatments)
F(CC)= (0.98)2=0.9604
f(Cc)= 2(0.98)(0.02)= 0.0392
f(cc)= (0.02)^2= 0.0004
freq at birth x fitness Freq Rep Norm
0.9604 x 1 0.9604 0.9608
0.0392 x 1 0.0392 0.0392
0.0004 x 0 0.0000 0.0000
Total—————> 0.9996

The new f(c) = 0.5 (0.0392)= 0.0196
Selection has almost NO effect on frequency of this allele
(due to the disease being very rare; not many homozygous individuals)
*rare recessive mutations: frequency should be stable
*if selection is not acting on allele, the frequency would stay the same)