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Archy [21]
3 years ago
8

. In centaurs, the allele for curly tails (T) is incompletely dominant to the allele for straight tails (t). The hybrid genotype

results in wavy tails. In a population of 1500 centaurs, 315 have curly tails, 820 have wavy tails, and the remainder has straight tails.
a. Determine the allele frequencies of: T = ____________ t = ____________

b. Determine the genotype frequencies of TT = ___________ tt = ____________



​
Biology
1 answer:
fiasKO [112]3 years ago
8 0

Answer:

a. Determine the allele frequencies of:  

  • f(T) = 0.485 ≅ 0.48
  • f(t) = 0.515 ≅ 0.52

b. Determine the genotype frequencies of the exposed generation

  • F (TT) = 0.21
  • F (Tt) = 0.55
  • F (tt) = 0.24

the genotype frequencies of the next generation

  • F(TT) = 0.23
  • F(Tt) = 0.5
  • F(tt) = 0.27

Explanation:

Due to technical problems, you will find the complete answer and explanation in the attached files

Download pdf
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If two organisms show a developmental homology you would also definently expect them to?
san4es73 [151]

If two organisms show a developmental homology you would also definitely expect them to share genetic homologies.

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7 0
1 year ago
The wechsler intelligence scale for children (wisc) is a good indicator of innate "genetic" ability for:
goblinko [34]

Answer:

The wechsler intelligence scale for children (WISC) is a good indicator of innate "genetic" ability for children of six to sixteen years old.

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Wechsler Intelligence Scale for Children (WISC) is basically a system of testing IQ of children from six to sixteen years old. This testing was created by David Wechsler, who stated that intelligence is a global quality and it should be reflected in children's verbal as well as non-verbal abilities.

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Point to remember:

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<em>Note:</em><em> This answer is very generic based on the statement of the question. If you have some specific options of the questions, please post it again with them or comment. Thank you.</em>



3 0
3 years ago
Indicate whether each of the following statements is true of depurination (DP), deamination (DA), or pyrimidine dimer formation
solniwko [45]

Answer:

- This process is caused by spontaneous hydrolysis of a glycosidic bond: depurination and deamination

- This process is induced by ultraviolet light:  pyrimidine dimer formation

- This can happen to guanine but not to cytosine: depurination

- This can happen to thymine but not to adenine:  pyrimidine dimer formation

- This can happen to thymine but not to cytosine: none

- Repair involves a DNA glycosylase: deamination

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- Repair involves DNA ligase: depurination, deamination and  pyrimidine dimer formation

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Explanation:

Depurination is the loss of purine bases (either adenine or guanine), while deamination refers to the removal of an amino group. During depurination, a β-N-glycosidic bond is cleaved by hydrolysis and a nucleic base is released (either adenine or guanine). All DNA bases may undergo deamination, except thymine (since thymine does not have an amino group). The ultraviolet (UV) radiation can cause thymine or cytosine to form dimers (e.g., pyrimidine dimers), being thymine dimers the most common lesion when DNA is exposed to UV light. Pyrimidine dimers may be repaired by different excision mechanisms, e.g., nucleotide excision repair, where the recognition of the DNA damage leads to the removal of the DNA fragment containing the lesion. DNA glycosylases are enzymes involved in the mechanism of base excision, these enzymes recognize and remove damaged bases by hydrolysis of the glycosidic bond, producing an abasic (apurinic and apyrimidinic) site. A DNA ligase enzyme covalently joins two DNA molecules by forming a phosphodiester bond, which is required during these processes.

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