Question

it in wombats. a female who is a carrier of the color blind allele
1 the phenotypes of their offspring are:
32
24
26
36
518 total
enotypic ratio to be?
periment:
:1 for any posibility
expected (e) \t(o-e) \t(o-e)² \t\frac{(o-e)²}{e}
29.5 \t2.5 \t \t
24.5 \t-0.5 \t \t
29.5 \t-3.5 \t \t
29.5 \t6.5 \t \t
\t \t$\chi² = \sigma \frac{(o-e)²}{e}$ \t
hypothesis?
codominant, sex - linked alleles: black (b) and orange (o). a
black male. they produce the following offspring:
ypic ratio to be?
ent:
ned (o) \t(o - e) \t(o - e)² \t\frac{(o - e)²}{e}
\t \t \t
\t \t \t
\t \t \t
\t \t \t

Explanation:

Step1: Calculate \((o - e)^2\) for each row

• Row 1: \(o - e = 2.5\), so \((o - e)^2 = 2.5^2 = 6.25\)
• Row 2: \(o - e = -0.5\), so \((o - e)^2 = (-0.5)^2 = 0.25\)
• Row 3: \(o - e = -3.5\), so \((o - e)^2 = (-3.5)^2 = 12.25\)
• Row 4: \(o - e = 6.5\), so \((o - e)^2 = 6.5^2 = 42.25\)

Step2: Calculate \(\frac{(o - e)^2}{e}\) for each row (assuming \(e = 29.5\) for first and third rows, \(e = 24.5\) for second and fourth rows)

• Row 1: \(\frac{6.25}{29.5} \approx 0.212\)
• Row 2: \(\frac{0.25}{24.5} \approx 0.010\)
• Row 3: \(\frac{12.25}{29.5} \approx 0.415\)
• Row 4: \(\frac{42.25}{24.5} \approx 1.724\)

Step3: Sum the values of \(\frac{(o - e)^2}{e}\)

\(\chi^2 = 0.212 + 0.010 + 0.415 + 1.724 \approx 2.361\)

Answer:

The calculated \(\chi^2\) value is approximately \(2.36\) (rounded to two decimal places).