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Helga [31]
3 years ago
14

In Morgan's testcross of a gray-bodied, long-winged heterozygous female Drosophila with a homozygous recessive black-bodied, ves

tigial-winged male, the following offspring were obtained: 965 gray body, long wing; 944 black body, vestigial wing; 206 gray body, vestigial wing; 185 black body, long wing. Focusing only on the recombinant classes (gray body, vestigial wing and black body, long wing), the numbers of offspring of each type are similar (206 and 185). What accounts for the similar number of offspring of each recombinant phenotype?
Biology
1 answer:
Elden [556K]3 years ago
8 0

Answer:

The similar number of offspring of each recombinant phenotype occurs as a result of Crossing over between chromosomes that are reciprocal, such that when a recombinant chromosome of one type is formed, the opposite recombinant of that same type is also formed.

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2.86 A ball is dropped from rest from the top of a cliff that is 24 m high. From ground level, a second ball is thrown straight
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Answer:

Distance Below the top = 6,02 m

Explanation:

To get the Final velocity of the first ball (that will be the intial velocity of the second) you need to solve the kinematic equation for Velocity:

(1) V_{1f} =V_{1O}  - gt

As the ball is dropped from rest V_{1O} = 0, so:

(2) V_{1f} = - gt

Note that the velocity is going to be negative as the ball is going down. To get the time it would take the ball to reach de base you can use the kinematic equation for position:

(3) h_{1} = h_{1O} + V_{1O} *t - \frac{1}{2} gt^{2}

We need the answer when h=0, and from the initial conditions V_{1O} = 0, h_{1O} = 24m, you get:

(4) 24m = \frac{1}{2} gt_{1f}^{2}

Solving for t, with 9,8\frac{m}{s^{2}} and :

(5) t_{1f} = \sqrt{\frac{24m*2}{g} } = 2,21 s

Replacing this time in (2), the final velocity is:

(6) V_{1f} = -21,66  \frac{m}{s}

So the initial velocity of ball 2 is equial to this but oppossite in direction so: V_{2O}= -V_{1f} = 21,66  \frac{m}{s}

The general position equation for ball 2 is (considering h_{2O} = 0 :

(7) h_{2} = V_{2O} *t - \frac{1}{2} gt^{2}

They cross paths when h_{1}=h_{2} so:

(8) h_{1O} - \frac{1}{2} gt^{2} = V_{2O} *t - \frac{1}{2} gt^{2}

Rearranging:

(9) t_{cross} =\frac{h_{1O}}{V_{2O} }

Replacing values:

t_{cross} = 1,108 s

To get the absolute position replace t_{cross} on equation (3) or (7):

h_{cross} = 17,98 m

To get it below the top of te cliff:

h_{cross, from above} = 24 m - 17,98 m

h_{cross, from above} = 6,02 m

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