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olchik [2.2K]
3 years ago
13

Why must males inherit colorblindness and hemophilia from their mothers?

Biology
1 answer:
zhuklara [117]3 years ago
8 0
Because chromosome X carries these genes which lead To catch these diseases.
males catch these disease more than females because they have 1. chromosome X but females has 2 .
so the Recessive trait appear in males more than females .
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A ____________ is a testable possible explanation, while a ___________ is an explanation that has been tested extensively. A) th
viva [34]

A hypothesis is a testable possible explanation, while a theory is an explanation that has been tested extensively.

If you need to learn more and remember about this stuff better, watch the Amoeba Sister's <em>Casual and Scientific Use of "Theory" and "Law"</em>. Trust me, it helps!

:)I hope I answered your question well!:)


4 0
3 years ago
Read 2 more answers
In your experiment, you found the following offspring only:Tall green: 626 offspring.Tall yellow: 313 offspring.Short green: 309
Karo-lina-s [1.5K]

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).

Answer:

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

       Tall plants with green pods.

   b.  F2 Gentotype:

  • 1/16 TTGG
  • 2/16 TTGg
  • 1/16 TTgg
  • 2/16 TtGG
  • 4/16 TtGg
  • 2/16 Ttgg
  • 1/16 ttGG
  • 2/16 ttGg
  • 1/6 ttgg

      F2 Phenotype:

  • 9/16 Tall plants with green pods (T-G-)
  • 3/16 Tall plants with yellow pods (T-gg)
  • 3/16 Short plants with green pods (ttG-)
  • 1/16 Short plants with yellow pods (ttgg)

Phenotypic ratio → 9:3:3:1

2)  T-G- → 702 individuals

     T-gg → 234 individuals

     ttG- → 234 individuals

     ttgg → 78 individuals

3)

  • Null Hypothesis: The population is under Hardy-Weinberg equilibrium. The alleles of this population assort independently.
  • Alternative Hypothesis: The population is not in equilibrium. Alleles do not assort independently.

4) a. There is enough evidence to <u>reject the null hypothesis</u>, 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

  • Tall and Green pods are the dominant traits,
  • Short and yellow pods are recessive traits.

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:

  • 1/16 TTGG
  • 2/16 TTGg
  • 1/16 TTgg
  • 2/16 TtGG
  • 4/16 TtGg
  • 2/16 Ttgg
  • 1/16 ttGG
  • 2/16 ttGg
  • 1/6 ttgg

      Phenotype:

  • 9/16 Tall plants with green pods (T-G-)
  • 3/16 Tall plants with yellow pods (T-gg)
  • 3/16 Short plants with green pods (ttG-)
  • 1/16 Short plants with yellow pods (ttgg)

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 individuals-------------X = 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)

  • Hypothesis: The population is under Hardy-Weinberg equilibrium. The alleles of this population assort independently.
  • Alternative Hypothesis: The population is not in equilibrium. Alleles do not assort independently

4)    F2 composed of:

  • Tall green: 626 offspring.
  • Tall yellow: 313 offspring.
  • Short green: 309 offspring.  

                       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 <u>reject the null hypothesis</u>, 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.    

5 0
3 years ago
What determines weather a metamorphic rock foliated or non-folited?
Kaylis [27]
Foliated rocks have bands, or layers( whatever you want to call it) unifoliated rocks don't have that I think.
7 0
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I need help :ppppppppp
Sloan [31]

Answer:

A and B is the answer

6 0
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Which of the following is the
marishachu [46]
**Its nitrate**
Nitrogen (N) + Oxygen (O) = nitrate
3 0
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