[SOLVED] estimate the mean (average) persistence of species with nonplanktonic larvae.

Researchers investigated whether differing modes of dispersal could explain differences in the longevity of fossil species in one taxon of marine snails, the family Volutidae. Some volute snails had planktonic larvae that could disperse over great distances on ocean currents. Other volute snails had nonplanktonic larvae, which developed directly into adults without a swimming stage. The dispersal of snails with nonplanktonic larvae was limited by the distance they can crawl as adults.

The researchers studied the distribution of volute snail fossils in outcrops of sedimentary rocks along North America’s Gulf Coast. The rocks formed during the early Paleogene period, between 66 and 37 million years ago, and contain many well-preserved snail fossils. Based on features of the snail’s shell, the researchers classified each fossil species as having planktonic or nonplanktonic larvae. Each bar in the figure shows how long one snail species persisted in the fossil record.

You can estimate quantitative data (fairly precisely) from a graph. The first step is to obtain a conversion factor by measuring along an axis that has a scale. In this case, 25 million years (from 60 to 35 million years ago on the x-axis) is represented by a certain distance in centimeters. Measure that distance on your screen or in your textbook. Then use that distance to generate a conversion factor (a ratio) of million years/cm.

To estimate the time period represented by a horizontal bar on this graph, measure the length of that bar in centimeters and multiply that measurement by the conversion factor. For example, if you calculated a conversion factor of 3.6 my/cm, a bar that measures 1.1 cm on the graph represents a persistence time of 1.1 cm × 3.6 my/cm = 4 million years.

Estimate the mean (average) persistence of species with nonplanktonic larvae.

1 million years

2.2 million years

5 million years

28 million years