Fine spines (s), smooth fruit (tu), and uniform fruit color (u) are three recessive traits in cucumbers whose genes are linked on the same chromosome. A cucumber plant heterozygous for all three traits is used in a testcross. The progeny from this testcross are:
S U Tu 2
s u Tu 70
S u Tu 21
s u tu 4
S U tu 82
s U tu 21
s U Tu 13
S u tu 17__
Total 230
a. Determine the order of these genes on the chromosome.
b. Calculate the map distances between the genes.
c. Determine the coefficient of coincidence and the interference among these genes.
d. Draw the chromosomes of the parents used in the testcross.

Answers

Answer 1

Answer and Explanation:

We have the number of descendants of each phenotype product of the tri-hybrid cross.

S U Tu 2 s u Tu 70 S u Tu 21 s u tu 4 S U tu 82 s U tu 21 s U Tu 13S u tu 17

The total number, N, of individuals is 230.

In a tri-hybrid cross, it can occur that the three genes assort independently or that two of them are linked and the thrid not, or that the three genes are linked. In this example, in particular, the three genes are linked on the same chromosome.

Knowing that the genes are linked, we can calculate genetic distances between them. First, we need to know their order in the chromosome, and to do so, we need to compare the genotypes of the parental gametes with the ones of the double recombinants. We can recognize the parental gametes in the descendants because their phenotypes are the most frequent, while the double recombinants are the less frequent. So:

Parental)

s u TU (70 individuals)S U tu (82 individuals)

Double recombinant)

S U Tu (2 individuals)s u tu (4 individuals)

Comparing them we will realize that between

s u TU (parental)

s u tu (double recombinant)

and

S U tu (Parental)

S U TU (double recombinant)

They only change in the position of the alleles TU/tu. This suggests that the position of the gene TU is in the middle of the other two genes, S and U, because in a double recombinant only the central gene changes position in the chromatid.

So, the order of the genes is:

---- S ---- TU -----U ----

In a scheme it would be like:

Chromosome 1:

---s---TU---u--- (Parental chromatid)

---s---tu---u--- (Double Recombinant chromatid)

Chromosome 2

---S---tu---U--- (Parental chromatid)

---S---TU---U--- (Double Recombinant chromatid)

Now we will call Region I to the area between S and TU and Region II to the area between TU and U.

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 S and TU genes, and P2 to the recombination frequency between TU and U.

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. So:

P1 = (R + DR) / N

P1 = (21+17+4+2)/230

P1 = 44/230

P1 = 0.191

P2= = (R + DR) / N

P2 = (21+13+4+2)/230

P1 = 40/230

P1 = 0.174

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

P1 + P2= Pt

0.191 + 0.174 = Pt

0.365=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. Now we must multiply each recombination frequency by 100 to get the genetic distance in map units:

GD1= P1 x 100 = 0.191 x 100 = 19.1 MU

GD2= P2 x 100 = 0.174 x 100 = 17.4 MU

GD3=Pt x 100 = 0.365 x 100 = 36.5 MU

To calculate the coefficient of coincidence, CC, we must use the next formula:

CC= observed double recombinant frequency/expected double recombinant frequency

Note:

observed double recombinant frequency=total number of observed double recombinant individuals/total number of individualsexpected double recombinant frequency: recombination frequency in region I x recombination frequency in region II.

CC= ((2 + 4)/230)/0.174x0.191

CC=(6/230)/0.0332

CC=0.7857

The coefficient of interference, I, is complementary with CC.

I = 1 - CC

I = 1 - 0.7857

I = 0.2143

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Here are some key evolutionary significance of bryophytes:

Adaptation to land: Bryophytes are considered some of the earliest land plants.

They were the first plants to transition from aquatic to terrestrial habitats, paving the way for the colonization of land by other plant groups.