Question:

Scorpions on the sand dunes in Syria in September 2022 have the age distribution as shown in Figure P. Scorpions can live to a maximum of 90 days. In all the figure panels, the x-axis represents age class and the y-axis represents number of individuals. Assuming no immigration or emigration, which one or more of the age distribution panels Q, R, S, T is/are possible 30 days later?

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In kinship problems, \emph{multiply} relatedness along each parent–offspring link and \emph{sum} over independent paths if multiple exist. Haplodiploidy boosts sister relatedness to \(0.75\), which then halves to an aunt–niece value of \(0.375\).
Updated On: Aug 26, 2025
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The Correct Option is A, B

Solution and Explanation

Step 1: Apply aging with a 90-day lifespan.
In 30 days, individuals advance by one age class; those presently in 61–90 days will exceed 90 and \emph{die}. With the “no immigration/emigration” clause and no information about births, the only deterministic change is: \[ \begin{aligned} \text{(1–30)d} &\Rightarrow \text{(31–60)d},
\text{(31–60)d} &\Rightarrow \text{(61–90)d},
\text{(61–90)d} &\Rightarrow \text{exit (0 individuals)}. \end{aligned} \] Hence, 30 days later we expect \emph{zero} in 1–30 d; counts in 31–60 d equal the previous 1–30 d bar; counts in 61–90 d equal the previous 31–60 d bar.

Step 2: Compare with candidate panels.
- Q and R still show many 1–30 d individuals \(\Rightarrow\) would require births/immigration (ruled out).
- S shows zeros in 31–60 d and nonzero 1–30 d \(\Rightarrow\) violates the deterministic aging shift.
- T shows 1–30 d = 0, 31–60 d equal to P’s 1–30 d bar, and 61–90 d equal to P’s 31–60 d bar \(\Rightarrow\) exactly the expected pattern. Final Answer:\quad \(\boxed{\text{(D) T}}\)
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