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Ymorist [56]
3 years ago
11

In garden peas the allele for tall plants () is dominant over the allele for dwarf plants() A true breeding tall plant is bred w

ith a true breeding dwarf plant to produce heterozygous tall plants. These Ft heterozygotes then self fertilize to yield F2 offspring. Complete the Punnett squar combination re for this monohybrid cross with the appropriate genotype of the Fi gametes and the F2 genotype/phenotype
A) Tt Tall
B) TT Tall
C) T
D) t
E) tt Dwart
Biology
1 answer:
marta [7]3 years ago
5 0

Answer: This is called Monohybrid Experiment

Explanation: Monohybrid Cross

P Generation TT     *   tt

  Tall plants     Dwarf plants

F1 Generation  

T T

t Tt Tt

t Tt Tt

In F1 generation;

There are 100% Tt Genotype and 100% Tall plant Phenotype

F2 Generation

F1   *   F1

( Tt   *   Tt )

T t

T TT Tt

t Tt tt

In F2 Generation;

Genotype

There are 25% TT (homozygous dominant, tall plants).

50% Tt (heterozygous tall plants), and

25% tt (homozygous recessive dwarf plants).

This is how dwarf characteristics reappear in the second generation.

The Phenotype of F2 generation is 3:1  (Tall : Dwarf)

I have attached a document to this answer to facilitate effective understanding if there is anormalities in arrangement the Punnet Square.

Download docx
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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

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Total R1 alleles = 2*42+86 = 170

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Expected number of each phenotype:

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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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