Suppose that a dominant allele (P) codes for a polka-dot tail and a recessive allele (p) codes for a solid colored tail. If two heterozygous individuals with polka-dot tails mate, what's the probability of the phenotype and genotype combinations of the offspring?A. 50 percent chance of a polka-dot tail and 50 percent chance of a solid colored tail (50 percent PP; 50 percent pp)
B. 75 percent chance of a polka-dot tail and 25 percent chance of a solid colored tail (75 percent Pp; 25 percent pp)
C. 75 percent chance of a polka-dot tail and 25 percent chance of a solid colored tail (25 percent PP; 50 percent Pp; 25
percent pp)
D. 25 percent chance of a polka-dot tail and 75 percent chance of a solid colored tail (25 percent PP; 50 percent Pp; 25
percent pp)

Answers

Answer 1
Answer: Make a punnett square. The individuals are heretozygus (different) so their genotype is Pp.
P p
_______
P | PP | Pp |
_______
p | Pp | pp |
_______

There is...
Genotype (genetics)
25% chance homozygus polkadot. (PP)
25% chance homozygus solid (pp)
50% chance heterozygus polkadot. (Pp)

Phenotype (physical trait)
75% chance polkadot
25% chance solid.

The answer: C


*sorry the punnett square is so weird looking :) i hope this was helpful.



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Answer: The fill in the blank can be filled with Stressor.

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A grocer stacks oranges in a four-sided pyramid that is 6 layers high. Solve for the number of oranges in the pile.A. 36
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C. 105
D. 91

Answers

Each layer contains the square of the layer number.
So,
Let's figure out :
1 squared = 1 

2 squared = 4 
3 squared = 9 
4 squared = 16 
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6 squared = 36
Hence,
The total number of oranges in the pile = 1+4+9+16+25+36
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The answer is 91.

The question is based on on a mathematical topic called ''Geometry'', and is simply asking for the total number of the oranges in the four-sided pyramid.

Further Explanation

In geometry, a pyramid is a polyhedron that is constructed when the polygonal base is connected to an iconic surface. The base edge and apex form a triangle that is called a lateral face. A pyramid can also be with an n-sided base that has n + 1 vertices, n + 1 faces, and 2n edges.

But the first thing to do is to get the number of oranges in each phases of the four-sided pyramid. To do this, we square the root the numbers of the vertices.

1 squared = 1 [for the first phase of the pyramid]

2 squared   =4 [we got 4 as our answer for the second layer of the pyramid]

3 squared = 9 [9, for the third phase of pyramid]

4 squared = 16  

5 squared = 25  

6 squared = 36 [and 36 is the last in the pyramid]

The addition of the total oranges in each phase of the six layers is 91.

The total number of oranges in the pile = 1+4+9+16+25+36=91

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