1. Using the equations Pm' = Pf ' and Pf' = (1/2)(Pm + Pf), the following table of allele frequencies can be constructed for male and female cats at each generation, where:

XA = yellow allele

XB = black allele

Punnett squares can be constructed to tell the genotypic frequencies in the first and second generations (G1, G2):

Punnett Square for G1

Punnett Square for G2

G1 genotpyes = XAXB      x       XAY     and

a) The female offspring of the original parents will have allelic frequencies XA = 0.50 and XB = 0.50. The genotypic frequencies are:

XAXB (calico female) = 1.00

XAXA (yellow female) = 0.00

XBXB (black female) = 0.00

Males will have the allelic frequencies XA = 1.00 and XB = 0.00. Their genotypic frequencies will be:

XAY (yellow male) = 1.00

XBY (black male) = 0.00

b) For G2, the females will show the allelic frequencies of XA = 0.75 and XB = 0.25. Their genotypic frequencies will be:

XAXB (calico female) = 0.50

XAXA (yellow female) = 0.50

XBXB (black female) = 0.00

Males will have the same allelic frequencies as the females in G1; XA = 0.50, XB = 0.50. Their genotypic frequencies will be:

XAY (yellow male) = 0.50

XBY (black male) = 0.50

d)  I couldn't get the allele frequency graphs to scan in correctly, so if you have questions, please come and ask me.  Sex-linked genes behave erratically when the population is very small (generations 1 and 2). Once more individuals enter the breeding population, however, there is a greater variety of genotypes, and the allele frequencies stabilize as they can appear in a number of genotypic combinations not possible in the first few generations.

2. a) Below is the inbreeding kinship diagram for Ptolemy V (labelled ‘E’).

Ptolemy V has five possible inbreeding paths:

E D C B A Z Y X E

E D Y Z A B C X E

E D Y X E

E D C X E

So, the inbreeding coefficient of Ptolemy V is calculated:

FE = [(1/2)^7 (1 + 0)] + [(1/2)^7 (1 + 0)] + [(1/2)^3 (1 + 0)] + [(1/2)^3 (1 + 0)]

        = 0.0078125 + 0.0078125 + 0.125 + 0.125

        = 0.265625

b) OK…that was the easy part. Now look at the inbreeding kinship diagram for Berenike IV (labelled ‘J’). Ptolemy V (now labelled ‘A’…sorry!) is up top, and we have to remember that he has a starting inbreeding coefficient of 0.265625. Unfortunately, he is not the only common ancestor we have to deal with. Paths can be traced with Ptolemy IX (‘D’) Cleopatra III (‘E’), Cleopatra Selene (‘F’), and Ptolemy X (‘G’) as the common ancestors. Before we can calculate the inbreeding coefficient for Berenike IV, we need to know the ancestors’ inbreeding coefficients. Ptolemy IX is not inbred, so his inbreeding coefficient is 0. The others work out to have the same inbreeding coefficients:

 

Cleopatra III (‘E’)

FE = [(1/2)^3 (1 + 0.265625)]

        = 0.158203125

Cleopatra Selene (‘F’)

FF = [(1/2)^4 (1 + 0.265625)] + [(1/2)^4 (1 + 0.265625)]

        = 0.158203125

Ptolemy X (‘G’)

FG = [(1/2)^4 (1 + 0.265625)] + [(1/2)^4 (1 + 0.265625)]

        = 0.158203125

So now we can start to work on Berenike IV. The total number of paths is (drum roll, please)……………………………………. 14:

J H F E B A D G I J

J H F E C A D G I J

J H G E B A D F I J

J H G E C A D F I J

J H F D A C E G I J

J H F D A B E G I J

J H G D A B E F I J

J H G D A C E F I J

J H G I J

J H F I J

J H F E G I J

J H G E F I J

J H G D F I J

Since many of the paths are the same length and have the same common ancestor, we can boil the inbreeding coefficient calculation for Berenike IV down to:

FJ = 8 [(1/2)^8 (0.265625 + 1)] + 2 [(1/2)^3 (0.158203125 + 1)] +
            2 [(1/2)^5 (0.158203125 + 1)] + 2 [(1/2)^5 (0+ 1)]

        = 0.039550781 + 0.289550781 + 0.072387695 + 0.0625

        = 0.463989257