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Valentin [98]
4 years ago
13

Question 9 (2 points)

Biology
1 answer:
sergejj [24]4 years ago
4 0

Answer:

all of these affect enzyme activity

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Match these cell division phases with the appropriate processes:
Komok [63]

Answer:

1) Break down of the nuclear membrane allowing mitotic spindles to connect to kinetochores:A. Metaphase

2)APC degrades securin which allows separase to become active which degrades the cohesin rings: E. Anaphase

3) Dephosphorylation of nuclear pore and lamins: F.Telophase

4) cleavage of plasma membrane by actin and myosin contractile ring:D.Cytokinesis

5) copying of the genome:C. Interphase

6) formation of the metaphase plate:B. Prometaphase

Explanation:

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3 years ago
Suresh and Gail are students in a lab working on mice with defects in the function of the HOX3A gene. In the special mouse line
Furkat [3]

Answer:

C) Both Suresh and Gail could be correct.

Explanation:

The pre-messenger RNA (pre-mRNA) transcript obtained after transcription of eukaryotic genes must undergo several processing events, including a process known as intron splicing, where introns (i.e., the non-coding sections of an RNA transcript) are removed and exons (coding regions) are joined to form a mature mRNA molecule. The three prime untranslated (3′-UTR) region of this mRNA can also bind to regulatory non-coding RNAs such as, for example, miRNAs which inhibit gene expression by inhibiting translation and/or by triggering its degradation. Moreover, the 3′-UTR region may also contain silencer sequences that bind to repressors in order to inhibit gene expression. On the other hand, translation refers to the process by which an ordered polypeptide chain (i.e., a protein) is synthesized by using the information contained in an mRNA molecule. In consequence, in the case under consideration, the mutation in the second gene could affect both RNA processing and the regulation of translation, thereby equally affecting HOX3A protein synthesis.

4 0
3 years ago
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.    

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