So if a couple is going to have 6kids what is the probabilit

So if a couple is going to have 6-kids, what is the probability of having 3 girls and 3 boys? What it then the probability of NOT having 3 girls and 3 boys? Now make a bar graph with the 6 different outcomes along the X-axis, and probability along the y-axis. The main difference when considering an autosomal condition is that you likely will have to take into account the dominance and recessive-ness of the trait that may alter the 1/2 to % probability. For instance, consider that in dogs, the condition of having black fur (B) is completely dominant to having white fur. In this case the white fur allele (b) is a mutant allele in the pathway that leads to the deposition of pigment in fur - in a heterozygote (Bb) the one \'B\' allele leads to wildtype pigment deposition and black fur color. Therefore, If two heterozygous black dogs were mated, what would be the probability of: A litter of 5 pups, four with black fur, then 1 with white fur, in this specified order. A litter of 5 pups, four with black fur and 1 with white fur, any order. In humans, eye color is complex but sometimes boils down to a simple dominant to recessive relationship among alleles. So consider the brown eye color allele \'B\' which deposits melanin in the iris of the eye to yield brown pigmented eyes. Interestingly, it is related to the gene that deposits melanin in the skin; more on that later. The allele \'b\' is a recessive mutant allele where the pathway to melanin deposition in the iris is disrupted (the \'B\' protein a either not produced at all, or the protein that is coded for has an active site that can\'t function) so that \'bb\' individuals have eyes that lack melanin and appear blue. If heterozygous children produce children, what are the following probabilities? The first two children will have blue eyes? That among a total of four children, 2 will have blue eyes, and 2 will have brown eyes? The first child will have blue eyes, but then the next three will have brown eyes?

Solution

Answer:

8).

Bb x Bb

BB (Black): Bb (Black) : bb(white) = 1 : 2 : 1

Black : white = 3 : 1

Black fur dog ratio = 3/4 (dominant)

white fur dog ratio = 1/4 (recessive)

a). P(bbbbw) = P(bbbb) + P(w) = (3/4 *3/4 *3/4 *3/4 ) + 1/4 = (81/256) + 1/4 =145 /256

b). P(4b1w in any order) = P(bbbbw) + P(bbbwb) + P(bbwbb)+P(bwbbb)+P(wbbbbb) = 81/1024 + 81/1024 +81/1024 +81/1024 +81/1024 = 405/1024