"GENETICS: BINOMIAL PROBABILITY, DOMINANT/RECESSIVE LETHAL TRAIT. The following example uses binomial probability for predicting the likelihood of a normal child in a family of 4 children. The calculated variable ODDS1 is probability of one normal child in family of N_CHILD children.The calculated variable PROBNORM is the probability of having N_NORMAL normal children. In the current example, the gene for disability is dominant (DOMINANT=1). Only the recessive genotype qq is normal.If the recessive genotypeis disabling, then DOMINANT=0 should be used. Select either 1 or 0 for the Dadand Mom's genetic makeup. gene #1 gene #2 QQ 1 1 Enter DGENE#1 and DGENE#2 for Dad. Qq 1 0 Enter MGENE#1 and MGENE#2 for Mom. qQ 0 1 qq 0 0 (c) PCSCC, Inc., 1993 *** Answer to Problem *** Set DGENE#1=1, DGENE#2=0, DOMINANT=1 (Q means disability), MGENE#1=1, MGENE#2=0,N_CHILD=4, N_NORMAL=2 (2 normal in 4). The probability of 2 normal children in 4 is PROBNORM=0.211 or about 21% for this event. Type any key to exit. ||A man and women each carry a disabling dominant gene Q and their pedigrees indicate that they are both heterozygous Qq. (a) If such a couple marry and have four children what is the probability of two being normal and two having the disability? Type comma key to see answers. Type (F2) to return to application file."
