Two codominant alleles, lm and ln, determine the human mn blood type. suppose that the lm allele occurs with a frequency of 0.80 in a population of eskimos on a small arctic island. match the expected frequencies to the m, mn, and n blood types in the population on the island for two mating scenarios: if random mating occurs, and if the inbreeding coefficient for this population is 0.05. mm frequency mn frequency nn frequency answer bank mating scenario random mating inbreeding coefficient of 0.05 0.040 0.640 0.648 0.320 0.048 0.304

Explanation:

Let p be the frequency of LM Allele and q is the frequency of LN Allele

Two alleles are Co-dominant

and p=0.8

so q=1-p=1-0.8=0.2

and inbreeding coefficient or F=0.05 (it is the degree to which two alleles are more likely to be homozygous rather than heterozygous)

The frequency of Homozygotes LMLM = (1-F) p2 +Fp

=p2-Fp2+Fp

=p2+Fp(1-p)

=p2+Fpq

Therefore, The frequency of Homozygotes LMLM in the Eskimo population will be

f(LMLM)= (0.8)2 + 0.05 (0.8) (0.2)

=0.648

Similarly The frequency of Homozygotes LNLN = q2+Fpq

or f(LNLN) = (0.2)2+0.05 (0.8) (0.2)

=0.048

And The frequency of Heterozygotes LMLN =2pq (1-F)

f(LMLN) = 2 (0.8)(0.2) (1-0.05)

=0.304

The expected frequencies of different blood types are

f(MM)=0.648, f(MN)=0.304, f(NN)=0.048

enzyme active site and substrate specificity

enzymes bind with chemical reactants called substrates. there may be one or more substrates for each type of enzyme, depending on the particular chemical reaction. in some reactions, a single-reactant substrate is broken down into multiple products