please explain in detail with answers You are mapping three linked genes in toma

Question

please explain in detail with answers

You are mapping three linked genes in tomato plants. The recessive alleles for the three genes (when in the homozygous condition) cause absence of anthocyanin pigment (a), hairless plants (h), and jointless fruit stems (j), respectively. Two true-breeding parents were crossed, producing trihybrid (heterozygous) F1s, which were then test-crossed. The 3000 offspring were grouped into the following phenotypic classes (a trait not mentioned in a particular phenotypic class can be considered to be dominant or wild-type for that trait):

              Phenotype                                                   Number Observed

              hairless                                                        249

              jointless, hairless                                         40

              jointless                                                       931

              wild type for all three                                  270

              anthocyaninless, jointless, hairless              278

              anthocyaninless, hairless                             941

              anthocyaninless                                            32

              anthocyaninless, jointless                            259

a) What are the original parental phenotypes?

b) Which of the three genes is in the middle?

c) Calculate the map distance between the three genes

Explanation / Answer

True breeding parents were crossed.

They are AAHHJJ and aahhjj.

The off spring will be AaHhJj

The off spring was test crossed. (Test cross is cross with recessive pure line)

So, the cross is between AaHhJj and aahhjj.

From AaHhJj parent, there will be following gametes.

AHJ

AHj

AhJ

Ahj

aHJ

aHj

ahJ

ahj

If there is no linkage, all the above gametes will be same proportion.

From aahhjj parent all gametes will be ahj type

If there is no linkage,

AaHhJj,

AaHhjj

AahhJj

Aahhjj

aaHhJj

aaHhjj

aahhJj

aahhjj

will be in same proportion. It means each one of them should be in 3000/8 = 375 in number.

Here is the table showing the number of off spring resulted in the cross. I have modified table to show all phenotypes.

Phenotype

Number Observed

anthocyanin, hairless, jointed

249

anthocyanin, hairless, jointless

40

anthocyanin, hairy, jointless

931

Wild type for all three

270

anthocyaninless, hairless, jointless

278

anthocyaninless, hairless, jointed

941

anthocyaninless, hairy, jointed

32

anthocyaninless, hairy, jointless

259

First, we have to identify parental classes:

If we see observed frequency, in the table, we would understand that distribution is equal. So, there is linkage.

If we observe the table carefully, off springs with “anthocyanin, hairy, jointless phenotype” and “anthocyaninless, hairless, jointed phenotype” are in 931 and 941 in number, respectively. This is very high compared to other phenotypes. These are parental combinations. Therefore they are present in high frequency.

Their genotypes will be AaHhjj and aahhJj, respectively.

Please notice that they are reciprocals of each other and have opposite phenotypes.

In AaHhJj parent, one chromosome contains AHj genes and its homologous chromosome contains ahJ alleles.

Next, we have to identify double cross overs:

As per the table, “anthocyanin, hairless, jointless phenotype” and “anthocyaninless, hairy, jointed phenotype” are 40, and 32 in number, respectively. This is very low compared to other phenotypes. They are double cross overs.

Their genotypes will be Aahhjj and aaHhJj, respectively.

Please notice that they are reciprocals of each other and have opposite phenotypes.

Now, we have to find out the middle gene.

AaHhjj and aahhJj are parental.

Aahhjj and aaHhJj are double crossovers.

In double cross overs, cross over occurs two times. Once between first and second genes and next between second and third genes. So, only gene, which recombines in double cross over is the middle gene on the linear chromosome.

Find out the difference between parentals and double cross overs.

The only difference is gene H.

Parent AaHhjj is similar to double cross over AahhJj except for H gene.

Parent aahhJj is similar to double cross over aaHhJj except for H gene.

So, H gene must be the middle gene.

So, the order of genes is AHJ.

Now I am adding genotypes to the table. Please note that alleles coming from recessive parent are same all progeny. So, I am ignoring those alleles.

Phenotype

Genotype

Number Observed

anthocyanin, hairless, jointed

AhJ

249

anthocyanin, jointless,hairless

Ahj

40

anthocyanin, hairy, jointless

AHj

931

Wild type for all three

AHJ

270

anthocyaninless, jointless, hairless

Ahj

278

anthocyaninless, hairless, jointed

ahJ

941

anthocyaninless, hairy, jointed

aHJ

32

anthocyaninless, hairy, jointless

aHj

259

We have to find the cross overs between A and H loci, first.

AHj is parent type.

AhJ and Ahj are offsprings resulted due to cross over between A and H loci.

ahJ is also a parent type.

aHJ and aHj are offsprings resulted due to cross over between A and H loci.

So, AhJ, Ahj, aHJ and aHj are cross overs between A and H loci. They are all together 249+40+32+259 = 580.

Total number of off springs is 3000.

Cross over percentage between A and H loci is 580 x 100/3000 = 19.33%

Now we find the cross overs between H and J loci.

AHj is parent type.

AHJ and aHJ are off springs resulted due to cross over between H and J loci.

ahJ is also a parent type.

ahj and Ahj are off springs resulted due to cross over between H and J loci.

So, AHJ, aHJ, ahj and Ahj are cross overs between H and J loci. They are all together 270+32+278+40 = 620.

Total number of off springs is 3000.

Cross over percentage between A and H loci is 620 x 100/3000 = 20.67%

One percent cross over is equal to 1 map unit.

Questions:

a) What are the original parental phenotypes?

Parental genotypes are AaHhJj and aahhjj.

AaHhJj phenotype is anthocyanin present, hairy and jointed

and

aahhjj phenotype is anthocyanin absent, hairless and jointless

b) Which of the three genes is in the middle?

H gene must be the middle gene.

c) Calculate the map distance between the three genes

Map distance between A and H is 19.33 map units.

Map distance between H and J is 20.67 map units.

Map distance between A and J is approximately equal to summation of distances between A to H and H to J.

Phenotype

Number Observed

anthocyanin, hairless, jointed

249

anthocyanin, hairless, jointless

40

anthocyanin, hairy, jointless

931

Wild type for all three

270

anthocyaninless, hairless, jointless

278

anthocyaninless, hairless, jointed

941

anthocyaninless, hairy, jointed

32

anthocyaninless, hairy, jointless

259

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.