Question

Suppose you cross two inbred (purebred) strains of mice, black and white, that have different average diastolic arterial blood pressure values, 99 mm Hg (black) and 141 mm Hg (white). You obtain F1’s with an intermediate average value of 120 (ranging between 116 and 124). After interbreeding the F1’s to produce F2’s, you measure the diastolic arterial blood pressure of these F2’s. You find that only 1 out of 1100 mice shows a blood pressure like the original parental black mouse like (99). This experiment was performed so as to greatly control environmental conditions. What does this say about the genetic control of blood pressure in mice? Estimate, roughly, the number of loci which contribute to this quantitative trait in this experiment.

Answer 1

• The parental blood pressure was 99 and 141 mm Hg.
• But the F1 had intermediate blood pressure around 120 mm Hg.
• This means the blood pressure is a genetic phenomenon that appears as a blend of two genes; one from mother and one from father. However, in F2, only 1 out of 1100 have blood pressure like the original parents (99).
• Rest 1099 have different blood pressure. This very large variation is possible only if characters are governed by multiple genes, a quantitative phenomenon.

To determine the number of genes involved, we have a simple formula

n (no. of loci) = (1/4)^n

= 1/1100 = (1/4)^5 [ 4^5 is approximately 1000)

So there are 5 loci responsible for blood pressure.