Pedigree analysis suggests that ear holes in professor Gana are inherited as a s

Question

Pedigree analysis suggests that ear holes in professor Gana are inherited as a simple Mendelian dominant trait. Professor Gana's do not express the trait but her grandma did have ear holes. Professor Ganas' husband does not have the trait but one of their three biological children does. What are the following ?. The first two of Professor Gana's children not have ear holes Professor Ganas third child has ear hole point If this union produces 5 children, what is the probability that 3 will have ear holes and 2 will not?. What is the genetics phenomenon that describes why Professor Ganas' mother who has the dominant ear hole allele but does not express the phenotype? Assume that the dominant allele that causes ear holes has a penetrance value of 30%. In other words, 30% of heterozygotes carrying the allele will have ear holes. If a homozygous unaffected person has children with a heterozygote carrying this mutant allele, what is the probability that an offspring will have ear holes? Show detailed calculations. What is the probability of rolling two six-sided dice and obtaining the following numbers 1 or 5 = 1/12chance An even number on both dice 1/2 chance

Explanation / Answer

(a) energy stored= 1/2 c v^2

U=1/2 (5.5x10^-6)(15)^2

U=6.1875 x10^-4 j .

U= 618.75 uJ

(b) when the distance between the plated is doubled , the capacitance reduces to half and the potential difference doubles

so new energy= U'= (1/2)(C/2)(2v)^2

U'= 2[(1/2)c V^2]

U'= 2 U = 2 times initial energy

U'= 2(6.1875 x10^-4 j .)

U'=1.2375 x10^-3 j .

U'= 1237.5 uj

(c) when the battery is inserted back, the volatge is same as in part 1 but the capacitace is half.

so C=2.75 uF

energy stored= (1/2)(2.75x10^-6)(15)^2

energy stored= 3.094x10^-4 j

energy stored=309.4uj

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