Question

1. with a scalpel, strip a small, thin, transparent layer of cells from the inside of a fresh onion leaf. 2. place it gently on a clean, dry slide, and add a drop of iodine solution (or methylene blue). cover with a coverslip. 3. observe under the microscope. 4. locate the cell wall and the nucleus. label figure 2.10. 5. count the number of onion cells that line up end to end in a single line across the diameter of the high-power (40×) field. ____ based on what you learned in section 2.4 about measuring diameter of field, what is the high-power diameter of field (hpd) in micrometers? ____ μm

Explanation:

To solve for the high - power diameter of field (HPD), we need to recall the relationship between the low - power diameter of field (LPD) and the high - power diameter of field. Usually, the low - power objective (e.g., 10×) has a known diameter of field. Let's assume the low - power diameter of field (LPD) is 1400 μm (a common value for a 10× objective). The magnification of the low - power (LP) is \(M_{LP}=10\times\) and the magnification of the high - power (HP) is \(M_{HP} = 40\times\).

Step 1: Recall the formula for the diameter of the field of view

The formula that relates the diameter of the field of view (\(d\)) and the magnification (\(M\)) is \(d_1\times M_1=d_2\times M_2\), where \(d_1\) and \(M_1\) are the diameter and magnification of the first objective, and \(d_2\) and \(M_2\) are the diameter and magnification of the second objective.

If we know the low - power diameter of field (\(d_{LP}\)) and the magnifications of low - power (\(M_{LP}\)) and high - power (\(M_{HP}\)), we can solve for the high - power diameter of field (\(d_{HP}\)) using the formula \(d_{HP}=\frac{d_{LP}\times M_{LP}}{M_{HP}}\)

Step 2: Substitute the values

Let's assume \(d_{LP} = 1400\space\mu m\), \(M_{LP}=10\times\) and \(M_{HP}=40\times\)

\[d_{HP}=\frac{1400\space\mu m\times10}{40}\]

First, calculate the numerator: \(1400\times10 = 14000\)

Then, divide by the denominator: \(\frac{14000}{40}=350\space\mu m\)

(Note: If the number of onion cells that line up end - to - end across the high - power field is \(n\), and we know the low - power diameter of field \(d_{LP}\) and the number of cells \(n_{LP}\) across the low - power field, we can also use the formula \(d_{HP}=\frac{d_{LP}\times n_{LP}}{n_{HP}}\). But since the number of cells across the high - power field is not given in the problem statement, we use the magnification - based formula. If we assume a typical low - power diameter of field of 1400 μm for a 10× objective, the high - power (40×) diameter of field is 350 μm)

Answer:

If we use the magnification - based approach with a typical low - power diameter of field of 1400 μm (for 10× magnification) and high - power magnification of 40×, the high - power diameter of field is \(\boldsymbol{350}\) μm. (If the number of cells is provided, we can use the cell - counting method. For example, if \(n\) cells are counted across the high - power field and we know the low - power diameter of field \(D\) and the number of cells \(N\) across the low - power field, \(HPD=\frac{D\times N}{n}\))