Genetically Viable Populations Flashcards
Long term genetic health. Populations must be large enough to:
A) Retain reproductive fitness by avoiding ID
B) Retain evolutionary potential
C) Avoid accumulation of new deleterious alleles
Retaining reproductive fitness
Total avoidance of inbreeding depression is impossible, therefore how much inbreeding can be tolerated without significant ID? Soule (1980) Ne = 50 to avoid inbreeding depression. Many captive population N= less than 50, they are too small.
Retaining evolutionary potential
Franklin (1980) stated Ne of 500 was minimum. Balance between loss of drift and gain by mutation. Lande (1995) upped it, Ne = 5000. Only 10 % of mutations are useful.
Retaining evolutionary potential caution
A minimum Ne of 500 or 5000 must be treated with caution.
1) These minimum sizes ignore natural selection
2) Fitness is a priority for evolutionary potential but these estimates are derived from peripheral characters.
3) These estimates consider heterozygosity, but ignores allelic diversity.
A serviere IB will occur at these sizes.
Species with an Ne of less than 500
Depletion of genetic diversity, reduced ability to evolve to novel threats, inbreeding, IB. Species will require more an increasing conservation effort.
Avoiding accumulation of new deleterious alleles
In large populations deleterious alleles kept at low frequencies. In small populations mildly deleterious alleles become ‘neutral’.
Chances are that IB is more likeyly to cause extinction than ‘mutation meltdown’.
Genetic goals in species conservation
Very few programmes have genetic goals. Target Ne can be used to measure success. If captive species need an Ne of 500+ then more space is needed. Only solution is to compromise, biologists must lower the viability threshold to accommodate more species. We can manipulate Ne/N ratios in captive populations to maximise Ne. In most captive populations Ne = 0.3.
