Question

Two autosomal genes control horn color in dragons. Pure-breedinggold-horned dragons were mated to pure-breeding silver-horneddragons. All of the F1 were gold. The F1 were intermated and the F2generation consisted of 147 gold, 17 silver and 92 bronze.
Fill in the blanks with whole numbers to indicate the geneticallybased phenotypic ratio that should be hypothesized to explain theF2 data.
gold: silver: bronze
Conduct a chi-square test to test the appropriate type ofepistasis.
In the chi-square test,
The expected number of gold-horned dragons is
The expected number of silver-horned dragons is
The expected number of bronze-horned dragons is

Answer 1

Answer:

The genetically based phenotypic  ratio for of Gold: Silver: Bronze = 9:1:6

The expected number of gold-horned dragons is = 144

The expected number of silver-horned dragons is = 16

The expected number of bronze-horned dragons is = 96

Explanation:

Given that; F2 generation consisted of:

147 gold

17 silver

92 bronze.

Total autosomal genes control horn color in dragons = 256

When we conducted the punnet square for the Pure-breedinggold-horned dragons were mated with the  pure-breeding silver-horneddragons, then intercrossed them for the F2 generation, It is seen that;

The genetically based phenotypic  ratio for of Gold: Silver: Bronze = 9:1:6

∴

The expected  number of gold-horned dragon = $256*\frac{9}{16}$

= 144

The expected number of silver- horned dragon = $256*\frac{1}{16}$

= 16

The expected number of Bronze-horned dragon = $256*\frac{6}{16}$

= 96

Null Hypothesis = The observed ratio of Gold:Siver:Bronze

=144:16:96

=9:1:6

Observed          Expected               (OF - EF)²                 $\frac{(OF-EF)^2}{EF}$

Frequency         Frequency

(OF)                    (EF)

147                       144                           9                                0.062500                    

17                         16                              1                                0.062500

92                       96                             16                               0.166667

Total:                                                                                      0.291667

Since Degree of Freedom = 2

Our P-Value = 0.864

∴ Our P. Value is very high that we accept the null hypothesis