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Banana plants, which are triploid, are seedless and therefore sterile. Propose a possible explanation.

Short Answer

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In bananas, the homologous chromosomes cannot pair up during synapsis of meiosis as they are triploid. As a result, meiosis is affected, and viable gametes are not produced.

Thus, a zygote with a triploid number of chromosomes cannot be produced, and therefore, the bananas are sterile.

Step by step solution

01

Meaning of Ploidy

The number of chromosome sets present in a cell is referred to as ploidy.In other words, it is the number of chromosomes present in the cell nucleus.

The chromosomes usually occur in pairs and are described as called diploids.However, cells with a single set of chromosomes are called haploids, such as gametes cells.

Similarly, cells with three sets of chromosomes are called triploids. Thus, polyploids are cells that have three or more three sets of chromosomes.

02

Meiosis

Meiosis is cell division that occurs in diploid cells. It involves two nuclear divisions that produce haploid daughter cells. These cells possess one set of chromosomes, which means the chromosome number is reduced to half.

During meiosis, the chromosome set is split to produce gametes or germ cells with half the number of chromosomes as the parent cell. As a result, haploid gametes cells are produced.

The gametes fertilize to produce a zygote that gives rise to a diploid organism.

03

Triploid bananas are sterile

In diploid organisms, the homologous chromosomes pair up during synapsis of prophase I.However, bananas have an extra set of chromosomes. This chromosome set with no homologous set fails to pair up during synapsis.

Thus, meiosis is disrupted, and no viable gametes are produced. These gametes cannot form a zygote with triploid sets of chromosomes. As a result, the bananas are seedless and sterile.

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Most popular questions from this chapter

Propose a possible reason that the first naturally occurring mutant fruit fly Morgan saw involved a gene on a sex chromosome and was found a male.

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 鈥渄egree of freedom鈥 (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?

Gene A, B, and C are located on the same chromosome. Test crosses show that the recombination frequency between A and B is 28% and that between A and C is 12%. Can you determine the linear order of these genes?

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For each type of offspring of the test-cross in Figure 15.9, explain the relationship between its phenotype and the alleles contributed by the female parent. (It will be useful to draw out the chromosomes of each fly and follow alleles throughout the cross.)

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