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Oduvanchick [21]
3 years ago
15

A student encounters an animal embryo at the eight-cell stage. The four smaller cells that comprise one hemisphere of the embryo

seem to be rotated 45 degrees and to lie in the grooves between larger, underlying cells (i.e., spiral cleavage).
This embryo may potentially develop into a(n):___________.
A) turtle.
B) earthworm.
C) sea star.
D) fish.
E) sea urchin.
Biology
1 answer:
frutty [35]3 years ago
4 0
The embryo may develop into an C.) earthworm
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B

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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

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