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Pavlova-9 [17]
2 years ago
7

The question refers to the following table, which compares the percent sequence homology of four different parts (two introns an

d two exons) of a gene that is found in five different eukaryotic species. Each part is numbered to indicate its distance from the promoter (for example, Intron I is the one closest to the promoter). The data reported for species A were obtained by comparing DNA from one member of species A to another member of species A.
Regarding these sequence homology data, the principle of maximum parsimony would be applicable in ________. (A) distinguishing introns from exons (B) determining degree of sequence homology (C) selecting appropriate genes for comparison among species (D) inferring evolutionary relatedness from the number of sequence differences
Biology
1 answer:
White raven [17]2 years ago
7 0

The question refers to the following table, which compares the percent sequence homology of four different parts (two introns and two exons) of a gene that is found in five different eukaryotic species. Each part is numbered to indicate its distance from the promoter (for example, Intron I is the one closest to the promoter). The data reported for species A were obtained by comparing DNA from one member of species A to another member of species A.

Regarding these sequence homology data, The principle of maximum parsimony would be applicable in Inferring evolutionary relatedness from the number of sequence differences.

What is principle of maximum parsimony?
Maximum A phylogenetic t
ree can be inferred using the character-based method of parsimony by reducing the total number of evolutionary steps needed to explain a given set of data assigned to the leaves.

Generally speaking, the parsimony principle states that it is preferable to use the simplest explanation possible to explain the data. Parsimony in phylogenetic analysis refers to the notion that a relationship hypothesis that calls for the fewest character changes is more likely to be accurate.

Therefore,

The principle of maximum parsimony would be applicable in Inferring evolutionary relatedness from the number of sequence differences.

To learn more about principle of maximum parsimony from the given link:

brainly.com/question/28427948

#SPJ4

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A cross is made between homozygous wild-type female Drosophila (a^+ a^+ b^+ b^+ c^+ c^+) and triple-mutant males (aa bb cc) (the
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Answer:

a is the middle gene.

Distance [b-a]= 24.7 mu

Distance [a-c]= 15.8 mu

Distance [b-a} = 40.5 mu

Explanation:

A homozygous wild-type female drosophila (a⁺b⁺c⁺/a⁺b⁺c⁺) is crossed with a homozygous recessive male (abc/abc). <u>The order of the genes here is arbitrary. </u>

The F1 is heterozygous for the three genes (a⁺b⁺c⁺/abc). The F1 females were test crossed (crossed with abc/abc males).

The F2 shows the following phenotypic ratios:

  • 320 a⁺b⁺c⁺
  • 308 a b c
  • 102 a⁺ b c⁺
  • 112 a b⁺ c
  • 66  a⁺ b⁺ c
  • 59 a b c⁺
  • 18 a⁺ b c
  • 15 a b⁺ c⁺

Total = 1000

The male parent is homozygous recessive for the 3 genes, so the observed phenotypes of the offspring correspond to the gametes received from the mother.

Recombination during meiosis is a rare event, so the most abundant gametes are always the parentals:  a⁺b⁺c⁺ and abc.

The least abundant gametes, following the same logic, are the double crossovers (DCO): a⁺bc and ab⁺c⁺.

<h3><u>1st. Determine the gene order</u></h3>

Compare the parental and the DCO gametes. The allele that is switched corresponds to the middle gene. In this case, gene a is in the middle of the other two.

<h3><u>2nd Determine the single crossover gametes</u></h3>

The F1 mother that generated all 8 types of gametes had the genotype b⁺a⁺c⁺/bac (correct order of genes).

  • The single crossover (SCO) gametes resulting from recombination between genes b and a are b⁺ac and ba⁺c⁺.
  • The single crossover (SCO) gametes resulting from recombination between genes a and c are b⁺a⁺c and bac⁺.
<h3><u>3) Calculate the recombination frequencies between genes </u></h3>

Recombination frequency (RF) = #Recombinants/Total progeny

  • RF [b-a]= (102+112+18+15)/1000= 0.247
  • RF [f-br]= (66+59+18+15)/1000= 0.158
<h3><u /></h3><h3><u>4) Calculate the distance in map units </u></h3>

Distance (mu) = RF x 100

Distance [b-a]= 0.247 × 100 = 24.7 mu

Distance [a-c]= 0.158 × 100 = 15.8 mu

Distance [b-a} = 24.7 mu + 15.8 mu = 40.5 mu

<h3><u>The gene map therefore looks like: </u></h3>

b------------24.7 mu--------------------------a---------15.8 mu-----------c

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