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ollegr [7]
2 years ago
12

Gene A has two alleles (A1 and A2). A1A1 homozygotes are twice as likely to survive from birth to reproductive age as heterozygo

tes or A2A2 homozygotes (w11=1, w12=0.5, w22=0.5). A population starts with 40 individuals of each genotype allowed to mate at random. Their progeny then survive according to the fitness values given above. What will be the allele frequencies in those progeny when they reach reproductive age
A. 0.40 A1, 0.60 A2 B. 0.50 A1, 0.50 A2 C. 0.60 A1, 0.40 A2 D. 0.80 A1, 0.20 A2 E. All A1
Biology
1 answer:
Usimov [2.4K]2 years ago
8 0

Answer:

The correct answer is C: <em>0.60 </em><em>A1</em> and <em>0.40 A2</em> will be the allele frequencies in those progeny when they reach reproductive age

Explanation:

<em>Genotype</em>                                <u><em>A1A1 </em></u><em>  </em><u><em>A1A2</em></u><em> </em><u><em>A2A2 </em></u>

<em>Relative aptitude, w</em>                  1           0.5    0.5

<em>Number of individuals</em>                 40            40     40

<em>Initial allelic frequency</em>      p0= (40+20)/120=0.5    q0= (40+20)/120=0.5

<em>Zygote frequency</em>                  p2= 0.25 2pq=0.25 q2=0.625

<em>Relative contribution</em>         0.25x1=0.25     0.5x0.5=0.25 0.25x0.5=0.125

<em>of each genotype</em>              

<em>Average aptitude W</em>              W= 0.025 + 0.25 + 0.125 = 0.625

<em>Population</em>                    AA= 0.25/0.625    AB=0.25/0.625  BB=0.125/0.625

<em>Genotype frequency</em>             AA= 0.4             AB=0.4          BB=0.2  

<em>New Allelic frequency</em>       p1=0.4+(0.4/2)=0.6   q1=0.2+(0.4/2)=0.4

  • <em>Total number of individuals</em>: 120
  • <em>Initial allelic frequency</em>:

                  (number of homozygote individuals + half number of  

                   heterozygote individuals) / Total number of individuals

  • <em>Relative contribution of each genotype</em>:

                           Zygote frequency x Relative aptitude

  • <em>Average aptitude W</em>: It is the sum of relative contribution of each genotype to the next generation.  

                    wA1A1 x p2 + WA1A2 x 2 x p x q + WA2A2 x q2

  • <em>Population Genotype frequency</em>:

                Relative contribution of each genotype / Average aptitude

  • <em>Allelic frequency</em>:

                      Homozygote population genotype frequency +  half  

                           heterozygote population genotype frequency

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The correct answer is - C. genus and species.

The final scientific name of an animal represents two things, its genus, and its species.

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Since the name Canus reminds me a lot of Canis, I will take the wolf as an example:

Kingdom: Animalia

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2 years ago
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How would signaling be affected if a mutation caused a g protein to replace gdp with gtp on its own without needing to be activa
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Answer:

This mutation will produce a conformational change capable of maintaining the receptor continuously in its activated mode

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Which of the following claims about the TYR, TRP2, and TRP1 mammalian genes is most likely to be accurate?
Illusion [34]

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The option a is correct.

Explanation:

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Tyrosinase enzyme is important for the synthesis of melanin, eye pigments and hair colour. The synthesis of all these is completed in three distinct reactions  catalysed by TRP1, TRP2 and Tyr genes. These work as operon and the protein product is almost 40% similar of the three genes.

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What situation best represents how DNA sequencing, PCR, and gene probes might be used together? Multiple Choice Scientists might
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Answer:

Scientists might replicate a strand of DNA using PCR before sequencing it. Once the sequence is known, they can produce a corresponding gene probe

Explanation:

PCR refers to the polymerase chain reaction that amplifies the small sample of DNA into multiple copies in three steps. These steps are denaturation of sample DNA to produce single-stranded template strand, binding of primer to the template and elongation. The multiple copies of the sample DNA are then used to decipher its sequence using various sequencing methods.

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2 years ago
In a flowering plant, tall (T) is dominant to short (t), and blue flowers (B) is dominant to white flowers (b). A tall plant wit
Bumek [7]
<h3>Answer: </h3><h3 />

25%

<h3>Explanation:</h3>

Following a cross between <u>Ttbb</u> and <u>ttBb</u>, the resulting offspring would be:

TtbB , TtbB , Ttbb , Ttbb , ttbB , ttbB , ttbb , ttbb , TtbB , TtbB , Ttbb , Ttbb , ttbB , ttbB , ttbb , ttbb

To have a clearer idea, see the attached image.

Of the 16 possible combinations from this dihybrid cross, there will be:

4 tall plants with blue flowers

4 tall plants with white flowers

4 short plants with blue flowers

4 short plants with white flowers

There are 4 possible genotypic recombinations that would yield short plants with white flowers out of a possible 16, so the chance that the offspring will be short with white flowers would be 4/16 which is 1/4 or 25%

<h3>Hope this helps</h3>

6 0
2 years ago
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