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Mademuasel [1]
3 years ago
6

In certain mice, long tails (T) are dominant over short tails (t). If 2 hybrid mice mate, what will the outcome be?

Biology
1 answer:
yuradex [85]3 years ago
7 0

Answer:

3 long tails : 1 short tail

Explanation:

This question involves a single gene coding for tail length in mice. The allele for long tail (T) is dominant over the allele for short tail (t). This means that an heterozygous mice will possess the long tail length.

According to this question, in a cross between two hybrid or heterozygote mice i.e. Tt × Tt, the following gametes will be produced by each parent:

Tt - T and t

Using these gametes in a punnet square (see attached image), the following will be produced: TT, Tt, Tt and tt.

Offsprings with genotype TT, Tt and Tt will have a LONG TAIL while genotype  tt will have a SHORT TAIL. Hence, the phenotypic ratio will be 3 long tails : 1 short tail.

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if a cell has 24 chromosomes, how many chromosomes would each of its four daughter cells have after meiosis?
andrew-mc [135]

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3 years ago
A virus is a tiny, infectious particle that can reproduce only by infecting a host cell. ... Nor do viruses have cells: they're
Llana [10]

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6 0
3 years ago
Two balloons having the same charge of 1.2 × 10-6 coulombs each are kept 5.0 × 10-1 meters apart. What is the magnitude of the r
Pavel [41]
The answer is <span>5.2 × 10-2 newtons.

To calculate this, we will use Coulomb's Law:
</span>F = \frac{k *  Q_{1} *  Q_{2} }{ r^{2} }
where F is repulcive force, k is constant, Q is charge, r is distance between charges.
<span>
If:
</span>Q_{1} = Q_{2} = 1.2 × 10^{-6} C²
r = 5.0 × 10^{-1} C²
k = 5.0 × 10^{9} \frac{N*m^{2} }{ C^{2} }

Then:
<span>F = \frac{5.0* 10^{-1} \frac{N* m^{2} }{ C^{2}} * 1.2 * 10^{-6} C * 1.2 * 10^{-6} C }{(5.0* 10^{-1} m)^{2}}
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3 years ago
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