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Natali [406]
3 years ago
11

You are a botanist looking at the plant color in a new breed of Snapdragon. The two parent plants are Blue and Green and when yo

u cross them you make F1 plants that are Cyan in color (a light aqua color). The ratio of offspring is as follows: There were 42 plants with Blue flowers (R1), 39 plants with Green flowers (R2) and 86 plants with Cyan flowers. Is the number of your progeny consistent with a single-gene incomplete dominance inheritance pattern
Biology
1 answer:
irakobra [83]3 years ago
4 0

Answer:

Incomplete dominance is the inheritance pattern where the dominant allele did not mask the recessive allele completely and form a mix of both alleles. Here the inheritance is the incomplete inheritance. The ratio of F2 generation is 1:2:1.

Given:

R1R1 = 42

R2R2 = 39

R1R2 = 86

Total R1 alleles = 2*42+86 = 170

Total R2 alleles = 2*39+86 = 164

Total alleles = 334

Frequency of allele R1 = 170/334 = 0.51

Frequency of allele R2 = 164 / 334 = 0.49

Expected number of each phenotype:

Total population = 167

Blue = R1R1 = 0.51 * 0.51 * 167 = 43.44

Green = R2R2 = 0.49 * 0.49 * 167 = 40.10

Cyan = 2*R1*R2 = 2*0.51*0.49*167 = 83.46

Phenotype     Observed(O)    Expected (E)    O-E      (O-E)2     (O-E)2/E

Blue                 42                   43.44               -1.44       2.0736     0.0477

cyan                86                   83.46                 2.54     6.4516     0.0773

green              39                    40.1                    -1.1       1.2100      0.0302

Total              167                     167                                               0.1552

Chi-square value = 0.155

Degrees of freedom = no. of phenotypes – 1

Df = 3-1 = 2

Critical value = 5.99

Chi-square value of 0.155 is less than the critical value of 5.99. So we accept the null hypothesis.

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Volgvan

Answer:

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Explanation:

We know the up-thrust on the fish, U = weight of water displaced = weight of fish + weight of air in air sacs.

So ρVg = ρ'V'g + ρ'V"g where ρ = density of water = 1 g/cm³, V = volume of water displaced, g = acceleration due to gravity, ρ'= density of fish = 1.18 g/cm³, V' = initial volume of fish, ρ"= density of air = 0.0012 g/cm³ and V" = volume of expanded air sac.

ρVg = ρ'V'g + ρ"V"g

ρV = ρ'V'g + ρ"V"

Its new body volume = volume of water displaced, V = V' + V"

ρ(V' + V") = ρ'V' + ρ"V"

ρV' + ρV" = ρ'V' + ρ"V"

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(ρ - ρ")V'  = (ρ' - ρ)V"

V'/V" = (ρ - ρ")/(ρ' - ρ)

= (1 g/cm³ - 0.0012 g/cm³)/(1.18 g/cm³ - 1 g/cm³)

= (0.9988 g/cm³ ÷ 0.18 g/cm³)

V'/V" = 5.55

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V' = V - V"

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