Question

FORKED LINE PROBABILITY:

You are analyzing three traits with the following offspring probabilities:

	Dominant Phenotype	Recessive Phenotype
Trait 1: Cystic fibrosis	90% Healthy	10% Diseased
Trait 2: Eye color	60% Brown	40% Blue
Trait 3: Blood Type	30% A , 20% B , 10% AB	40% O

What Proportion of offspring are expected (assume independent traits):

A) To be diseased w/ blue eyes + blood type A

B) To display the dominant phenotype for at least two traits?

C) To display the dominant phenotype for at least one trait?

NOTE:

USE Forkline method , and how you got to your answer.

Answer 1

Following is the forked line diagram for the probability of traits in question:

(The percent probabilities have been converted into fractions for easy understanding)

Calculations are as follows:

(A) From forked line diagram, the probability can be calculated by navigating through the lines and multiplying the individual probabilities (as the three events are independent).

Probability of offspring to be diseased= 1/10

Probability of offspring with blue eyes= 4/10

Probability of offspring with blood type A= 3/10

The probability of offspring who are diseased with blue eyes and blood type A= 1/10 × 4/10 × 3/10= 12/1000

Hence, the proportion of offspring who are diseased with blue eyes and blood type A= 12/1000 × 100= 1.2%

(B) To find out the proportion of offspring with at least two dominant phenotypes, we have to calculate the individual probabilities of offspring with two or more dominant phenotypes and then add them to obtain the total probability (because the events are mutually exclusive).

Following is the list of conditions when at least 2 dominant phenotypes are the outcome:

1) Healthy + Brown + blood type O = 9/10 × 6/10 × 4/10 = 0.216

2) Healthy + blue + blood type A = 9/10 × 4/10 × 3/10 = 0.108

3) Healthy + blue + blood type B = 9/10 × 4/10 × 2/10 = 0.072

4) Healthy + blue + blood type AB = 9/10 × 4/10 × 1/10 = 0.036

5) Healthy + Brown + blood type A = 9/10 × 6/10 × 3/10 = 0.162

6) Healthy + Brown + blood type B = 9/10 × 6/10 × 2/10 = 0.108

7) Healthy + Brown + blood type AB = 9/10 × 6/10 × 1/10 = 0.054

8) Diseased + Brown + blood type A = 1/10 × 6/10 × 3/10 = 0.018

9) Diseased + Brown + blood type B = 1/10 × 6/10 × 2/10 = 0.012

10) Diseased + Brown + blood type AB = 1/10 × 6/10 × 1/10 = 0.006

Total probability= 0.216 + 0.108 + 0.072 + 0.036 + 0.162 + 0.108 + 0.054 + 0.018 + 0.012 + 0.006 = 0.792

So, proportion of offspring with at least two dominant phenotypes = 0.792 × 100 = 79.2%

(C) The proportion of offspring with at least one dominant phenotype can be calculated by subtracting the probability of offspring with all recessive phenotype from 1

Probability of all recessive phenotype (diseased + blue + blood type O) = 1/10 × 4/10 × 4/10 = 0.016

Probability of at least one dominant phenotype = 1-0.016 = 0.984

So, proportion of offspring with at least one dominant phenotype = 0.984 × 100 = 98.4%