the following statement is found in a recent report by ODFW
"It is emphasized that regardless of the stage in the life history, a decrease in survival is carried throughout the life cycle. For example, a 15% reduction in egg to emergent fry survival across an entire population would have the same impact as a 15% reduction in over-winter survival of juveniles, or a 15% reduction in marine survival rate, or a 15% increase in the mortality impact of fisheries. In addition, because survival is multiplicative, it is possible to use the results in Table 10 to gauge the net effect of changes in survival at several life stages at once. For example, a 30% decrease in freshwater survival due to habitat modifications on top of a 30% decrease in marine survival would result in an overall decrease in life cycle survival of [1 - (1 – 0.30)(1 – 0.30)] = 0.51 * 100% = 51% . The multiplicative feature of the survival relationship can be used to examine any number of possible combinations of changes from the current assumption about the future of survival through certain portions of the coho’s life history"

I was under the impression that a 15% increase in Egg to Emergent Fry mortality would have virtually no change in adult spawner abundance but a change in marine survival would have the greatest… did I miss something?

I'm no biologist, but it seems to be a simple math problem. I'm not being sarcastic, but I don't understand why one would have a greater impact to your final number than the other. Both sources of mortality should decrease the final population (returning spawners). The multiplicative impact is somewhat intuitive. But what do I know, I'm just a dumb engineer.

BB

Clerk,

It's true the way it's used in the example illustration. However, context matters. Salmon and steelhead populations are subject to both density dependent and density independent mortalities. The ODFW example is consistent with density independence.

A different outcome would be a coho or steelhead group that suffered a low egg to fry survival, but due to the low fry density, that population segment might have a higher than average fry to parr or smolt survival rate. Funny thing about fish populations; it isn't always simple math. If it was, we'd have right answers more often.

Sincerely,

Salmo g.