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Ratling [72]
2 years ago
10

In pointers dogs, the Punnett square for tail length is below (gene A). Use the Punnett Square below to determine the possible o

ffspring from a cross between two heterozygotes.
F1
A1 A2
A1 A1 A1 A1 A2
Never born Bow-tie tails
A2 A1 A2 A2 A2
Bow-tie tails Long tails
a. In terms of tail length what kind of inheritance does this gene show?
b. In terms of survival what kind of inheritance does this gene show?
Biology
1 answer:
kvv77 [185]2 years ago
5 0

Answer:

a) Autosomal dominant inheritance ⇒ The presence of at least one A1 allele is enough to express Bow-tie tails.

b) Autosomal recessive inheritance ⇒ The presence of at least one recessive allele A2 is enough for individuals to survive.

Explanation:

Due to technical problems, you will find the complete explanation in the attached files.

Download pdf
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When studying animal behavior, the distribution of organisms within a choice chamber can be studied to identify animal preferenc
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Answer:

c

Explanation:

Random distribution of the Isopods should be 50/50 in A and B.

Chi-square = \frac{(O - E)^2}{E} where O = observed frequency and E = expected frequency

Chamber         O                    E             Chi-square

  A                   18                   10             \frac{(18 - 10)^2}{10} = 6.4

  B                    2                    10             \frac{(2 - 10)^2}{10} = 6.4

<em>Total Chi-square </em>= 6.4 + 6.4 = 12.8

<em>Degree of freedom</em> = 2 - 1 = 1

<em>Tabulated Chi-square</em> (α = 0.05) = 3.8415

The calculated Chi-square exceeds the critical value, hence, it is significant and not due to random.

<em>The correct option is </em><em>c</em><em>.</em>

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Answer:

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Explanation:

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Explanation:

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