Answer:
Step-by-step explanation:
AaBBccDd x AabbCCDDd ---Parents
Aa x Aa = AA (1/4), Aa (1/2) & aa (1/4)
BB x bb = Bb (1)
cc x CC = Cc (1)
Dd x Dd = DD (1/4), Dd (1/2), dd (1/4)
a). The probability that any given offspring of this pair will be heterozygous at all loci = AaBbCcDd = ½ * 1*1*1/2 = ¼
b). The probability that an offspring will be homozygous at all four loci =AABBCCDD = ¼*0*0*1/4 = 0
c). The probability that an offspring will be phenotypically dominant = A_B_C_D_ = ¾ * 1 * 1* ¾ = 9/16