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Nitella [24]
3 years ago
7

A plant species has two possible gene variations for seed shape. One plant has smooth seeds, and another has wrinkled seeds. The

re is a no chance that their offspring will be born with wrinkled seeds. What can you conclude from this information?
The plant with smooth seeds is heterozygous for the dominant trait.




Both plants are heterozygous for smooth seeds.




The plant with the smooth seeds is homozygous for the dominant trait.




One parent is homozygous, and the other is heterozygous for the dominant trait.




Both genes are codominant in the offspring.
Physics
2 answers:
marusya05 [52]3 years ago
7 0
<span>The plant with the smooth seeds is homozygous for the dominant trait.</span>
vovikov84 [41]3 years ago
6 0

Answer: The correct answer is- The plant with the smooth seeds is homozygous for the dominant trait.

According to the Mendelian genetics, dominant trait is the one that expresses itself and masks the expression of the recessive trait. Dominant is depicted with capital letter and recessive with small letter.

Let us consider 'S' depicts the allele for smooth seed and 's' for the wrinkled seeds. There can be two genotypes for smooth seeds that is 'Ss' and 'SS' whereas only one genotype is possible for wrinkled ( 'ss') seeds.

As per the question, there is no chance of the offspring born with wrinkled seeds. This is only possible when the parental plant with smooth seeds is homozygous for the dominant trait ( like SS) . Therefore, only smooth trait will be visible in the offsprings ( with genotype Ss) of the parental plants and there will be no chance of wrinkled seeds (ss) in the offsprings.


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Hi there!

a)
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dB = Differential Magnetic field element

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

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i = Current in loop (0.460 A)

For a circular coil, the radius vector and the differential length vector are ALWAYS perpendicular. So, for their cross-product, since sin(90) = 1, we can disregard it.

dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{r^2}

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Taking out constants from the integral:
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B =\frac{\mu_0 i}{4\pi R^2}  \int\limits^{2\pi R}_0 \, ds

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b)

Now, we have an additional component of the magnetic field. Let's use Biot-Savart's Law again:
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In this case, we cannot disregard the cross-product. Using the angle between the differential length and radius vector 'θ' (in the diagram), we can represent the cross-product as cosθ. However, this would make integrating difficult. Using a right triangle, we can use the angle formed at the top 'φ', and represent this as sinφ.  

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r = \sqrt{z^2 + R^2 }\\\\sin\phi = \frac{R}{\sqrt{z^2 + R^2}}

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