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3 years ago

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Place the planets in order based on their average distance from the Sun, shown in the table in part B. Now, compare the sizes of

the planets. What relationship do you observe between a planet’s size and its proximity to the Sun

1 answer:

Jobisdone [24]3 years ago

3 0

Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune and then the possible Planet Nine

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In humans there is a dominant allele (A) for the absence of moles; while the recessive allele (a) results in the presence of mol

Art [367]

Answer:

a. $(p + q)^7$ = $p^7q^0+7p^6q+21p^5q^2+35p^4q^3+35p^3q^4+21p^2q^5+7pq^6+p^0q^7$

b. i. 0.75

ii. 0.000061

iii. 0.012

iv. 0.17

c. 0.67

Explanation:

a. The expansion of the binomial (p + q)7 would be such that:

$(p + q)^7$ = $p^7q^0+7p^6q+21p^5q^2+35p^4q^3+35p^3q^4+21p^2q^5+7pq^6+p^0q^7$

b. Both couples are heterozygous:

Aa x Aa

AA Aa Aa aa

Since A is dominant over a,

probability of having mole (aa) = 1/4

probability of not having moles = 3/4

<em>Therefore, the probability of the first child not having moles </em>= 3/4 or 0.75

ii. Let the probability of not having mole = p and the probability of having mole = q. From the binomial expansion:

$(p + q)^7$ = $p^7q^0+7p^6q+21p^5q^2+35p^4q^3+35p^3q^4+21p^2q^5+7pq^6+p^0q^7$

<em>Probability that all of the children will have moles </em>= $p^0q^7$

since p = 3/4 and q = 1/4

$p^0q^7$ = $(3/4)^0(1/4)^7$ = 0.000061

iii. <em>Probability that the first two children will have no moles and the last five will have moles</em> = $21p^2q5$

= $21(3/4)^2(1/4)^5$

= 0.012

iv. <em>Probability that 4 will have no moles and 3 will have moles out of the 7 children</em> = $35p^4q^3$

= $35(3/4)^4(1/4)^3$

= 0.17

c. <em>Probability that the child born without moles is a carrier of the a-allele = probability of heterozygou</em>s.

From the cross in (b), the genotypes of those born without moles are AA and 2Aa. Therefore, the probability of not having moles and be Aa is:

= 2/3 or 0.67

8 0

3 years ago