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castortr0y [4]
2 years ago
9

in roses, red is dominant over white. short stems are dominant over long stems. if you cross two heterozygous red, short-stemmed

roses, what are the chances of producing the desirable trait of long-stemmed red roses?
SAT
1 answer:
Galina-37 [17]2 years ago
3 0

Let

• R = red

• r = white

• S = short stem

• s = long stem

Then two heterozygous (RrSs) parents can each contribute one of the following allele pairs:

RS, Rs, rS, rs

so that the genotypes of their offspring have the following distribution:

\begin{array}{ccccc} & RS & Rs & rS & rs \\ RS & \boxed{RRSS} & \boxed{RRSs} & \boxed{RrSS} & \boxed{RrSs} \\ Rs & \boxed{RRSs} & RRss & \boxed{RrSs} & Rrss \\ rS & \boxed{RrSS} & \boxed{RrSs} & \boxed{rrSS} & \boxed{rrSs} \\ rs & \boxed{RrSs} & Rrss & \boxed{rrSs} & rrss \end{array}

The probability of producing offspring with either the SS or Ss genotype is then 12/16 = 3/4.

In other words, there are 4² = 16 possible genotypes among the offspring, and each parent has a 1/2 probability of contributing the S allele. There are 2 cases to consider:

• both parents contribute the S allele with probability (1/2)² = 1/4

• one parent contributes the S allele while the other contributes the s allele, and this can happen in 2 ways with probability 2 × (1/2)² = 1/2

These events are mutually exclusive, so the total probability is 1/4 + 1/2 = 3/4.

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Altitudes $\overline{ad}$ and $\overline{be}$ of acute triangle $abc$ intersect at point $h$. if $\angle ahb = 128^\circ$ and $\
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The Angle HCA is equal to 52°. This is arrived at using the knowledge of the Total value of Angles in a Triangle and the Total Value of Angles in a Polygon.

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<h3>What are the Steps to the Solution? </h3>

Step 1 - Recall that we have been given Angles AHB and BAH to be 128° and 28° respectively.

We also know that:

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Since A...therefore:

When ∠AHB (128°) and ∠BAH are taken from 180° we have DBA = ∠28°.

By observation, we can deduce that ∠BDE, ∠CDH, ∠CEH and ∠AEH are all right-angled triangles.


Using the above, we are able to repeat this process of solving for each angle until we have ∠HCA.

To verify that our answer is correct, recall that sum of angles in a polygon is 360°

That means:

∠BDA + ∠DHE + ∠CEH + ∠HCA = 360°

That is, 90+ 128 + 90 + 52 = 360°

See attached images and;

Learn more about Angles in a Triangle at:
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