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Chapter 15: Q15.4-1CC (page 309)

About 5% individuals with Down syndrome have a chromosomal translocation in which a third copy of chromosome 21 is attached to chromosome 14. If this translocation occurred in a parent’s gonad, how could it lead to Down syndrome in a child?

Short Answer

Expert verified

Chromosome 14-21 combines as a single chromosome in meiosis. It can combine with the normal chromosome leading to the condition of trisomy 21. It is a cause of Down syndrome.

Step by step solution

01

Description of Down syndrome

Down syndrome is otherwise known as chromosome 21 disorder. It is associated with conditions such as dental probes, endocrine problems, ear infections, and hearing problems. The development of body organs takes place in a delayed manner in this condition.

Thyroid and heart disease are other serious conditions that are associated with Down's syndrome.

02

Description of chromosomal translocation

The transfer of the segment of the chromosomes to other locations in the same chromosome is known as chromosomal translocation. The transfer can also take place to the other chromosomes. It can result in the formation of new chromosomes.

03

Translocation leading to Down syndrome

Trisomy is the condition of having an additional chromosome along with the normal chromosome. During the process of meiosis, the combination of chromosomes 14 and 21 takes place. The combined chromosome fuses with the normal chromosome resulting in the formation of trisomy 21.

Trisomy 21 is the genetic condition that leads to a defective condition known as Down's syndrome.

Hence, the translocation that occurs in the parent gene can result in Down syndrome in children.

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Most popular questions from this chapter

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?

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?

Crossing over is thought to be evolutionarily advantageous because it continually shuffles genetic alleles into novel combinations. Until recently, it was thought that the genes on the Y chromosome might degenerate because they lack homologous genes on the X chromosome with which to pair up prior to crossing over. However, when the Y chromosome was sequenced, eight large regions were found to be internally homologous to each other, and quite a few of the 78 genes represent duplicates. (Y chromosome researcher David Page has called it a "hall of mirrors.”). Explain what might be a benefit of these regions.

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?

Assume you are mapping genes A, B, C, and D in Drosophila. You know that these genes are linked on the same chromosome, and you determine the recombination frequencies between each pair of genes to be as follows: A-B, 8%; A-C, 28%; A-D, 25%; B-C, 20%; B-D, 33%.

  1. Describe how you determined the recombination frequency for each pair of genes.

  2. Draw a chromosome map based on your data.
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