Term 2 Lecture 13: Predicting the outcomes of a genetic cross with 2 or more alleles

Question 1

Multiplication rule

Answer 1

Chance of rolling a 4 on a dice two times in a row

1/6 x 1/6 = 1/36 probability

Question 2

Addition rule

Answer 2

Likelihood of rolling a 3 or a 4 when rolling a dice

1/6 + 1/6 = 1/3

Question 3

Binomial expansion

Answer 3

this is when transmission gets complicated
Binomial expansion is used to determine the probability of a complex situation with multiple possible outcomes
e.g. chance of a couple having a child with albinism followed by 2 subsequent children with pigmentation

(P+Q)ⁿ

P= probability of one event
Q= probability of alternate event
ⁿ = number of time the event occurs

Question 4

Binomial expansion example

Answer 4

What is the probability of 2 out of 5 of the couples children having albinism?

P= 1/4 (albinism is a recessive trait)
Q= 3/4 (1-P)
ⁿ = 5

(P+Q)⁵ =
P⁵ 5 albino
+
5P⁴Q 4 albino
+
10P³Q² 3 albino
+
10P²Q² 2 albino
+
5PQ⁴ 1 albino
+
Q⁵ no albino

Each of the terms in the expansion provides the probability of one particular combination of traits in the children

Probability that 2 of 5 are albino =
10P²Q³ = 0.26

Question 5

Observed rates of progeny may deviate from expected ratios by chance

Answer 5

• if ratios observed in actual genetic cross experiments don’t fit with Mendelian rules how do we know if it’s by chance?
• by using a chi square goodness of fit test
• indicates the probability that the difference between the observed and expected number of progeny is due to chance

Question 6

Chi square goodness of fit test

Answer 6

• statistical test that results in a probability
• starts with a null hypothesis

” There is no significant difference between the observed and expected number of progeny”
I.e. suggests any difference is due to error and not genetic cause

Chi square test allows us to accept or reject this hypothesis
However it can’t tell us whether results are right or wrong

Question 7

Chi square goodness of fit test example

Answer 7

P generation: purple and pink flowering plants

F1: all purple - so purple is Dom

F2: purple 105 pink 45

Expected 3:1 does it match up?

Phenotype: observed/expected
Purple: 105/ (3/4x150) = 112.5
Pink: 45 / (1/4x150) = 37.5

Chi square equation:

X² = (O-E)²/ E
= (105-112.5)²/112.5 + (45-37.5)²/37.5
= 56.25/112.5 + 56.25/37.5
= 0.5+1.5
Chi square value =2

Degrees of freedom = n-1 =2-1 = 1

Look up chi square value (2) in relation to degrees of freedom value (1) if P>0.05 there is not significant difference and you accept the null hypothesis.

Here P>0.05 so accept null

Question 8

Mendel’s model assumes

Answer 8

• alleles at one locus sort independently from alleles at another locus
• only one gene locus determines a trait
• alleles are either dominant or recessive
• genes only occur on nuclear chromosomes (actually can come from mitochondrial genome)

^ not everything fits the Mendel model, some genes are linked closely and can cross and segregate together (in prophase 1)

Question 9

How can we tell if alleles are linked?

Answer 9

F2 ratio will not be as expected
Due to crossovers occuring

Question 10

Transposons discovered in the 1940s

Answer 10

By Cold Spring Harbor Laboratory geneticist Barbara McClintock