Answer:
P (actually that randomly selected person has diabetes given as given by the lab) = 0.7725
Explanation:
A probability is a number, between 0 and 1, expressing the precise likelihood of an event happening.
Probability (person actually has diabetes) = 5.9%
P(lab test says that they have diabetes) = 0.0764
P (actually that randomly selected person has diabetes given as given by the lab) = 0.7725
Workings shown in the figure below.
Answer:
The correct answer would be 2 in 4.
According to the question Xo and XO show codominance and express themselves completely when present in heterozygous condition. Cats bearing XoXO show patchwork of black and orange fur and are called tortoiseshell cats.
Xo codes for orange color fur and XO codes for black color fur. In addition, Y chromosome does not contain any gene associated with fur color.
Now, genotype of mother cat is XOXO (orange fur). So, the gametes formed would be XO only.
The genotype of father cat is XoY(black fur). So, the gametes would be Xo and Y.
The cross would lead to the formation of two male cats each having XOY as their genotype and two female cats each with XOXo as their genotype.
Hence, both the male cats would show orange fur and both the female cats would show patchwork of orange and black fur.
Therefore, we can conclude that 2 out of 4 would exhibit tortoiseshell coloring.
