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vaieri [72.5K]
3 years ago
15

A short-tailed mutant of mouse was discovered. Multiple crosses of this mouse to normal mice produced 27 normal, long-talied mic

e and 25 short-tailed mice. A series of crosses among short tailed mice were made and and 21 short-tailed mice and 11 long-tailed mice were produced? Study these results and determine which phenotype is dominant and explain the ratios observed with regards to the genotypes of the parents in each cross.
Biology
1 answer:
poizon [28]3 years ago
4 0

Answer:

Allele for short tail is a dominant phenotype.

Explanation:

In the second generation , the cross between a group of short tailed mice  produces either short tailed or long tailed species.

This means the short tailed mice must be having the allele for long tail trait but did not expressed it.

Hence, when two alleles are presents and one is expressed while the other is suppressed then allele expressing its trait would be the dominant phenotype.

Hence, the allele for short tail is a dominant phenotype.

Cross between two short tailed mice will be as follows

Ss * Ss

SS, Ss, Ss, ss

Out of four three are short tailed while one "ss" is long tailed.

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Alu element at the PV92 locus on Chromosome 16. In this population, 436 people have a heterozygous genotype and 102 people have
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Answer:

Explanation:

According to the exercise we can infer that for the Alu insert:

p: positive allelic frequency

q: negative allelic frequency

maintaining that the population is in equilibrium we can carry out the following formula

p + q = 1 and pp + 2pq + qq = 1

looking for the genotype frequency we clear and obtain the following data

genotype frequency of 2pq = 436/1000 = 0.436

The genotype frequency of qq = 102/1000 = 0.102

This is how we look now:

number of positive people for Alu = 1000- (436 + 102) = 1000- 538 = 46

In this way it is resolved that:

genotypic frequency of pp = 462/1000 = 0.462

  p = 0.462 + (0.436 / 2) = 0.462 + 0.218 = 0.680

q = 0.102+ (0.436 / 2) = 0.102 + 0.218 = 0.320

According to the exercise carried out it is deduced:

The value of p in the population is = 0.68

We conclude that our prognosis showing a homozygous positive genotype is: 0.462

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