Conversation Biologist working with the Florida Panther have set out camera trap
ID: 149123 • Letter: C
Question
Conversation Biologist working with the Florida Panther have set out camera traps to estimate the overall population. Through a combination of live trapping and hair snares, the biologists have worked out the exact genetics for two alleles in all remaining individuals. The homozygous revessive state causes a debilitating disease and the panthers die before reproducing. Biologists have found a high incidence of heterzygous individuals. Help the Biologists determine the best conservation action in order to protect and boost the genetic health of the remaining population. The current population of of Florida Panthers in this example is 20 which is based upon available habitat. 18 of the Panthers are heterozygous [Aa]. Scientists have determined that 18 is the minimum viable population for the Florida panthers.
a. If eighteen of the panthers are heterozygous and if homozygous recessive individuals die before reproducing, what is the genotype for the remaining two panthers?
b. Using the Hardy – Weinberg equations, determine the frequencies of the dominant and recessive alleles.
c. This population of panthers reproduces and the resulting population is also twenty individuals. Based on the allele frequencies calculated in part b, calculate the genotypic frequencies for the:
homozygous dominant genotype –
heterozygous genotype –
homozygous recessive genotype –
d. How many of the twenty panthers in this generation will suffer from the disease and die?
e. What is the number of survivable Florida panthers?
Explanation / Answer
1) AA, homozygous dominant
2) Freq(aa)=0
Freq(Aa)=18/20
Freq(AA)=2/20
freq(a)=0+0.9/2=.45
freq(A)=.1+.9/2=.55
c) genotype freq(AA)=.55*.55=.30
freq(aa)=.45*.45=.20
freq(Aa)=1-(.2+.3)=.5
d) 20%of 20=4 die
e) 16 survive
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