Question

On the planet Abby, there is an homozygous lethal condition known as Gumball. Individuals with this condition chew gum till their jaws hurt. A study of 1000 heterozygote Abbers (Gg) shows that 600 show the Gumball phenotype. Homozygous wild-type (gg) Abbers never show the phenotype. If we cross of a wild-type gg male with an affected female, what is the probability that an arbitrarily selected F1 will be wild-type?
A. 0.6
B. 0.0
C. 1.00
D. 0.70
E. 0.3

Answer 1

Answer:

D. 0.7 would be wild-type healthy

E. 0.3 would be affected Gg

Explanation:

Using the Punnett square we cross an homozygous male (gg), that is always healthy vs an heterozygous female (Gg). The result is 2 gg individuals, and 2 Gg individuals. Both gg individuals are healthy.

In a population of 1000 Gg individuals, 600 show the Gumball phenotype (unhealthy). It means that from the two heterozygous Gg (from the Punnett square), 1.2 show the unhealthy phenotype.

Again, from the Punnett square we have 4 individuals, that represent the 100%. 1.2 are unhealthy Gg, and represent 30% in F1.

This 30%, is equivalent to 0.3, that is the option E.

100 - 30 % = 70 %.  This is 0.7 for healthy wild-type, option D.