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zhannawk [14.2K]
3 years ago
11

In hamsters, there is a dominant allele (S) for smooth fur and a recessive allele (s) for spiky fur. A unique characteristic of

these alleles is that the heterozygote (Ss) has its own phenotype - rough fur. In this way, we can determind the number of heterozygotes in the population. In a population of 1000 hamsters, 390 have smooth fur (SS), 470 have rough fur (Ss) and 140 have spiky fur (ss).
What are the allele frequencies in this population?
Frequency of S =
Frequency of s =
In the next generation of 1000 hamsters, what are the expected genotype frequencies?
Frequency (SS) =
Frequency (Ss) =
Frequency (ss) =
What are the expected number of each genotype in the next 1000 hamsters?
Number of SS Number of Ss Number of ss Have the allele frequencies changed?
Is the population in Hardy-Weinberg equilibrium?
have the allele sequencie: nanged is the populati Weinberg equitoThe expected genotype frequency of Ss (2pq) in the next generation is expected to be:
a. 0.469
b. 0.390
c. 0.938
d. 469
The expected number of smooth fur hamsters (SS) in the next generation of 1000 hamsters is:
Is the population of hamsters in Hardy-Weinberg equilibrium?
Biology
1 answer:
Oksana_A [137]3 years ago
3 0

Answer:

<em>What are the allele frequencies in this population?</em>

  • Frequency of S = 0.625  
  • Frequency of s = 0.375

<em>In the next generation of 1000 hamsters, what are the expected genotype frequencies?</em>

  • Frequency (SS) = 0.39          
  • Frequency (Ss) = 0.47          
  • Frequency (ss) = 0.14  

<em>What are the expected number of each genotype in the next 1000 hamsters?</em>

  • Individuals SS =  390
  • Individuals Ss =  469
  • Individuals ss =  140

<em>Have the allele frequencies changed?</em> No, they are the same

<em>Is the population in Hardy-Weinberg equilibrium?</em> Yes, it is.  

<em>The expected genotype frequency of Ss (2pq) in the next generation is expected to be: </em>0.469 (option a)

<em>The expected number of smooth fur hamsters (SS) in the next generation of 1000 hamsters is</em>: 390

Explanation:

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

     

Download pdf
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Following the selfing of the F1 from the cross between the two parental strains, what proportion of the F2 individuals will phenotypically resemble either one of the two parental lines?

= 7.03% of F2 progeny will resemble parent 1

= 2.34% of F2 progeny will resemble parent 2.

Explanation:

Parent 1: AA; b/b; D/D; e/e; F/F

Parent 2: a/a; B/B; d/d; E/E; f/f

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knowing that the F1 progeny is selfed, the possible gamete outcomes are:

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At the A, B and D loci, the F1 are heterozygotes. Therefore, in the F2 progeny, for each locus, 3/4 of the progeny will have the dominant trait while 1/4 will have the recessive phenotype. Further, 1/2 of all F2 progeny will have the haplotype eF and the other half will be Ef.

<u>Proportion of progeny resembling Parent 1: </u>

Parent 1 has the Phenoytpe AbDeF. As previously discussed, 3/4th of all F2 progeny has the phenotype A, 1/4th of F2 progeny has the phenotype b, 3/4th of F2 progeny has the phenotype D and 1/2 of F progeny has the phenotype eF.

Thus, the proportion of progeny resembling Parent 1

= 3/4 x 1/4 x 3/4 x 1/2 = 0.0703125

= 7.03% of F2 progeny will resemble parent 1.

<u>Proportion of progeny resembling Parent 2</u>:

Parent 2 has the Phenoytpe aBdEf. As previously discussed, 1/4th of all F2 progeny has the phenotype a, 3/4th of F2 progeny has the phenotype B, 1/4th of F2 progeny has the phenotype d and 1/2 of F progeny has the phenotype Ef.

Thus, Proportion of progeny resembling parent 2

= 1/4 x 3/4 x 1/4 x 1/2 = 0.0234375

= 2.34% of F2 progeny will resemble parent 2.

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