In your experiment, you found the following offspring only:Tall green: 626 offspring. Tall yellow: 313 offspring. Short green: 309 offspring. Stay with your original hypothesis/assumption, test it using chi2 test and analyze the result (whether you want to reject the hypothesis and why).(6p)

Complete question:

1. You found a new species of garden pea and have decided to repeat Mendel’s experiments. You have obtained two true-breeding pea plants that are tall (T) with yellow pods (g), and short (t) with green pods (G), respectively. Capital letters indicate dominant traits.

a. What is the genotype and phenotype of the F1 generation? (2p)

b. You allowed self-fertilization in F1 generation and obtained F2 generation. Using Punnett square to obtain the phenotypes, genotypes and their respective ratios. (4p)  

2. If you obtained 1248 offspring in the F2 generation in Question 1, calculate how many offspring of each phenotypic class you would expect to have. (4p)

3. When you finish questions 1 and 2, under what assumption/hypothesis did you perform your analysis/calculation? Is there an alternative assumption/hypothesis, which may result in a different conclusion? If yes, what is the alternative assumption/hypothesis and what kind of offspring and in what numbers do you expect to obtain? (4p)    

4. You only need to answer A or B to get this 6p. Please indicate which question you are answering. In your experiment, you found the following offspring only: Tall green: 626 offspring. Tall yellow: 313 offspring. Short green: 309 offspring.  

A: If you stay with your original hypothesis/assumption, test it using chi2 test and analyze the result (whether you want to reject the hypothesis and why).

B: If you believe that the alternative hypothesis fits better with your data, test it using chi2 test and analyze the result (whether you want to accept the hypothesis and why).

1) a. 100% of the progeny will be dihybrid, TtGg, exhibiting the phenotype    

       Tall plants with green pods.

   b.  F2 Gentotype:

      F2 Phenotype:

Phenotypic ratio → 9:3:3:1

2)  T-G- → 702 individuals

     T-gg → 234 individuals

     ttG- → 234 individuals

     ttgg → 78 individuals

3)

4) a. There is enough evidence to reject the null hypothesis, meaning that the difference between the observed number of individuals and the expected ones is statistically significant. The population is not under equilibrium H-W. Alleles do not assort independently.  

Explanation:

1) 1st Cross: True-breeding tall (T) with yellow pods (g) pea plant with a short (t) with green pods (G) plant

Parentals)         TTgg         x         ttGG

Gametes) Tg, Tg, Tg, Tg          tG, tG, tG, tG

F1) 100% of the progeny will be dihybrid, TtGg, exhibiting the phenotype    

    Tall plants with green pods.

2nd Cross: F1 self-fertilization

Parentals)  TtGg     x     TtGg

Gametes) TG, Tg, tG, tg

                TG, Tg, tG, tg

Punnett square)    TG         Tg           tG          tg

                 TG      TTGG     TTGg      TtGG      TtGg

                 Tg       TTGg      TTgg      TtGg      Ttgg

                 tG        TtGG      TtGg       ttGG       ttGg

                  tg        TtGg       Ttgg       ttGg       ttgg

F2) Gentotype:

      Phenotype:

Phenotypic ratio → 9:3:3:1

2) The total number of individuals in the F2 is 1248.

16 1248 individuals   100% of the progeny

9 T-G- X = 702 individualsX = 56.25%

3 T-gg X = 234 individuals X = 18.75%

3 ttG- X = 234 individuals X = 18.75%

1 ttgg X = 78 individuals X = 6.25%

3)

4)    F2 composed of:

                       Tall/Green        Tall/Yellow         Short/Green       Short/Yellow

Observed              626                  313                        309                        0

Expected               702                  234                       234                       78

(Obs-Exp)²/Exp     8.23                  26.7                        24                        78

X² = Σ(Obs-Exp)²/Exp = 8.23 + 23.7 + 24 + 78 = 136.93 ≅ 137

Freedom degrees = genotypes - number of alleles = 9 - 4 = 5

Significance level, 5% = 0.05

Table value/ Critical value = 11.07

X² = 137

X² > Critica value

137 > 11.07

There is enough evidence to reject the null hypothesis, meaning that the difference between the observed number of individuals and the expected ones is statistically significant. The population is not under equilibrium H-W. Alleles do not assort independently.    

the atmospheric gas that forms a mild acid when dissolved in water is called carbon dioxide. the correct answer is a. 2.) chemical weathering would cause an inscription on a marble headstone to become harder and harder to read over time.