In the tomato, three genes are linked on the same chromosome. Tall is dominant to dwarf, skin that is smooth is dominant to skin that is peachy, and fruit with a normal tomato shape is dominant to oblate shape. A plant that is true-breeding for the dominant traits was crossed to a dwarf plant with peachy skin and oblate fruit. The F1 plants were then testcrossed to dwarf plants with peachy skin and oblate fruit. The following results were obtained:151 tall, smooth, normal33 tall, smooth, oblate11 tall, peach, oblate2 tall, peach, normal155 dwarf, peach, oblate29 dwarf, peach, normal12 dwarf, smooth, normal0 dwarf, smooth, oblateConstruct a genetic map that describes the order of these three genes and the distances between them.

Genetic Map + distances between genes

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   ║---6.4%---║16.3%---║

   ║22.7%║

Explanation:

Available data:

We can recognize the parental gametes in the descendants because their phenotypes are the most frequent,

while the double recombinants are the less frequent.

And simple recombinant gametes produced by the cross, which frequencies are intermediate.  

Comparing the parental and the double recombinant we will realize that they only change in the position of the alleles S. This suggests that the position of the gene skin texture is in the middle of the other two genes, because in a double recombinant only the central gene changes position in the chromatid.

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Now we will call Region I to the area between T and S and Region II to the area between S and N.

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      RI      RII

Once established the order of the genes we can calculate distances between them, and we will do it from the central gene to the genes on each side. First We will calculate the recombination frequencies, and we will do it by region. We will call P1 to the recombination frequency between T and S genes, and P2 to the recombination frequency between S and N.

P1 = (R + DR) / N

P2 = (R + DR)/ N

Where: R is the number of recombinants in each region, DR is the number of double recombinants in each region, and N is the total number of individuals. To calculate the recombination frequency, we have to know that 1% of recombinations = 1 map unit = 1cm. And that the maximum recombination frequency is always 50%.  

N = 151 + 33 + 11 + 2 + 155 + 29 + 12 + 0 = 393

So:

Now, to calculate the recombination frequency between the two extreme genes, A and B, we can just perform addition or a sum:

P1 + P2= Pt

0.064 + 0.163 = Pt

0.227 = Pt

The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU).  One centiMorgan (cM) equals one map unit (MU). The map unit is the distance between the pair of genes for which one of every 100 meiotic products results in a recombinant product.

Genetic Map

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   ║---6.4%---║16.3%---║

   ║22.7%║

i believe the answer is c

spiders belong to a group of arachnids.

arachnids have the following traits :

they have two body parts.

they have eight legs unlike insects which have 6 legs.

they have no wings.

these traits distinguish them from insects.

the information that antonio is doubting us the statement that spiders differences are as a result of being first to evolve.

this is because the evolution of one species is independent of the other species.