Problem 2.2 (20 points)Black and yellow pigment cells (melanophores and xanthophores) are responsible for patterns onthe skin of Zebra Fish. We denote the number of melanophores at timetbyM(t) and the numberof xanthophores at timetbyX(t). Suppose that 10% of melanophores become (differentiate into)xanthophores per unit time. Conversely, 40% of xanthophores become melanophores per unit time, and 10% of xanthophores die per unit time. Finally, 10 melanophores and 20 xanthophores areborn per unit time.1. DefineP(t)

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Please let me assume this to complete your question;

P(t) = M(t) ÷ X(t)

ANSWER:

P(t) is the ratio of the number of melanophores to the number of xanthophores per unit time.

Therefore P(t) = 1.55

Step-by-step explanation:

KEY NOTE:

I) 10% of melanophore converts to Xanthophore per unit time.

II) 40% of Xanthophore converts to melanophore per unit time.

III) 10 melanophore and 20 Xanthophore are born per unit time

IV) 10% of Xanthophore die per unit time.

STEP 1: Conversion rate to number of birth per unit time.

For M(t) 10% of 10 = 1

For X(t) 40% of 20 = 8

Therefore;

M(t) = 10 + 8 - 1 = 17

X(t) = 20 + 1 - 8 = 13

STEP 2:

10% of Xanthophore die per unit time.

10% of 20 = 2

Therefore X(t) is;

13 - 2 = 11

The total X(t) = 11

The total M(t) = 17

P(t) = M(t)/X(t) = 17÷ 11 = 1.55

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