Using the information from problem 4, scientists do a further testcross using a heterozygote for height and nose morphology. The offspring are tall upturned snout, 40; dwarf upturned snout, 9; dwarf downturned snout, 42; tall downturned snout, 9. Calculate the recombination frequency from these data, and then use your answer from problem 4 to determine the correct order of the three linked genes.

Short Answer

Expert verified

18% is the frequency of recombination between the genes for height tall (T) and an upturned snout (S).

The correct order of the three genes on the chromosome is T-A-S.

Step by step solution

01

Recombination frequency

Recombinant frequency measures the proportion of recombinant offspring obtained when parents with specific traits are crossed.It represents the chances of recombination that occurs between the two genes in a chromosome.

Thus, recombinant frequencies between the genes help to calculate the distance between the genes on the chromosome.

The recombinant frequency between two genes is calculated using the following formula:

\({\rm{Recombination frequency}}\,{\rm{ = }}\frac{{{\rm{Number of recombinants}}}}{{{\rm{Total number of offsprings}}}}\,{\rm{ \times 100}}\)

02

Explanation for the recombinant frequency between genes for height and nose morphology

Four types of offspring were obtained in the testcross using heterozygote for height and tall morphology. The testcross using heterozygote with tall and an upturned snout produced four types of offspring.

The tall upturned snout (40) and dwarf downturned snout (42) were wild-type offspring, while dwarf upturned snout (9) and tall downturned snout (9) were recombinant offspring.

The total number of offspring produced in the cross is \(40 + 9 + 42 + 9 = 100\)

The total number of recombinants is \(9 + 9 = 18\).

The recombination frequency is calculated by substituting the values into the following equation:

\(\begin{aligned}{l}{\rm{Recombination frequency }}\,{\rm{ = }}\frac{{18}}{{100}}\,\,{\rm{ \times 100}}\\{\rm{Recombination frequency}}\,\, = \,\,18\% \,\end{aligned}\)

Thus,the recombination frequency for the genes between tall (T) and an upturned snout (A) is 18%.

03

To determine the order of the genes on the chromosome  

From question four, the recombination frequency for the genes between tall (T) and antennae (A) obtained was 12%. The recombination frequency for the genes between antennae (A) and an upturned snout obtained was 5%. While the recombination frequency for the genes between tall (T) and an upturned snout (A) is 18%.

From the recombinant frequencies obtained from three crosses, it is inferred that the distance between the genes for height (T) and nose morphology (S) is the highest, that is, 18%. In comparison, the recombinant frequency between the genes for antennae (A) and nose morphology (S) is the least, that is, 5%.

The data indicate that genes for antennae and nose morphology are located nearer than the genes for height and nose morphology. This exhibits that the order of the genes on the chromosomes must beT-A-S.

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Most popular questions from this chapter

Which one of Mendel’s laws describes the inheritance of alleles for a single character? Which law relates to the inheritance of alleles for two characters in a dihybrid cross?

Mitochondrial genes are critical to the energy metabolism of cells, but mitochondrial disorders caused by mutations in these genes are generally not lethal. Why not?

The\({\chi ^2}\)value means nothing on its own- it is used to find the probability that, assuming the hypothesis is true, the observed data set could have resulted from random fluctuations. A low probability suggests that the observed data are consistent with the hypothesis, and thus the hypothesis should be rejected, A standard cutoff point used by biologists is a probability of 0.05(5%). If the probability corresponding to the\({\chi ^2}\)value is 0.05or considered statistically significant, the hypothesis (that the genes are unlinked) should be rejected. If the probability is above 0.05, the results are not statistically significant: the observed data are consistent with the hypothesis.

To find the probability, locate your\({\chi ^2}\)value in the\({\chi ^2}\)Distribution table in Appendix F. The “degree of freedom” (pdf) of your data set is the number of categories (here,4 phenotypes), minus 1, so df=3.

(a). Determines which values on the df =3 line of the table your calculated\({\chi ^2}\)value lies between.

(b). The column headings for these values show the probability range for your\({\chi ^2}\)number. Based on whether there is non-significant (p\( \le \)0.05) or significant (p>0.05) difference between the observed and expected values, are the data consistent with the hypothesis that the two genes are unlinked and assorting independently, or is there enough evidence to reject this hypothesis?

Review the description of meiosis (see Figure 13.8) and Mendel’s laws of segregation and independent assortment (see Concept 14.1). What is the physical basis for each of Mendel’s laws?

Two genes of a flower, one controlling blue (B) versus white (b) petals and the other controlling round (R) versus oval (r) stamens, are linked and are 10 map units apart. You cross a homozygous blue oval plant with a homozygous white round plant. The resulting F1 progeny are crossed with homozygous white oval plants, and 1,000 offspring plants are obtained. How many plants of each of the four phenotypes do you expect?

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