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Chapter 7: Problem 18

Indicate the expected number of Barr bodies in interphase cells of individuals with Klinefelter syndrome; Turner syndrome; and karyotypes \(47, \mathrm{XYY}, 47, \mathrm{XXX},\) and \(48, \mathrm{XXXX}\)

Short Answer

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Answer: The expected number of Barr bodies for each condition are: 1. Klinefelter syndrome (XXY): 1 Barr body 2. Turner syndrome (XO): 0 Barr bodies 3. 47, XYY: 0 Barr bodies 4. 47, XXX: 2 Barr bodies 5. 48, XXXX: 3 Barr bodies

Step by step solution

01

Klinefelter syndrome

Klinefelter syndrome (XXY) has two X chromosomes and one Y chromosome. Using our formula, Barr bodies = 2 (number of X chromosomes) - 1 = 1 Barr body.
02

Turner syndrome

Turner syndrome (XO) has one X chromosome and no Y chromosome. Using our formula, Barr bodies = 1 (number of X chromosomes) - 1 = 0 Barr bodies.
03

47, XYY

47, XYY has one X chromosome and two Y chromosomes. Using our formula, Barr bodies = 1 (number of X chromosomes) - 1 = 0 Barr bodies.
04

47, XXX

47, XXX has three X chromosomes and no Y chromosome. Using our formula, Barr bodies = 3 (number of X chromosomes) - 1 = 2 Barr bodies.
05

48, XXXX

48, XXXX has four X chromosomes and no Y chromosome. Using our formula, Barr bodies = 4 (number of X chromosomes) - 1 = 3 Barr bodies. In conclusion, the expected number of Barr bodies for each condition is: 1. Klinefelter syndrome (XXY): 1 Barr body 2. Turner syndrome (XO): 0 Barr bodies 3. 47, XYY: 0 Barr bodies 4. 47, XXX: 2 Barr bodies 5. 48, XXXX: 3 Barr bodies

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Most popular questions from this chapter

How does \(X\) chromosome dosage compensation in Drosophila differ from that process in humans?

In a number of organisms, including Drosophila and butterflies, genes that alter the sex ratio have been described. In the pest species Musca domesticus (the house fly), Aedes aegypti (the mosquito that is the vector for yellow fever), and Culex pipiens (the mosquito vector for filariasis and some viral dis- eases), scientists are especially interested in such genes. Sex in Culex is determined by a single gene pair, \(M m\) being male and \(m m\) being female. Males homozygous for the recessive gene \(d d\) never produce many female offspring. The \(d d\) combination in males causes fragmentation of the \(m\) -bearing dyad during the first meiotic division, hence its failure to complete spermatogenesis. (a) Account for this sex-ratio distortion by drawing labeled chromosome arrangements in primary and secondary spermatocytes for each of the following genotypes: \(M m D d\) and \(M m d d .\) How do meiotic products differ between \(D d\) and \(d d\) genotypes? Note that the diploid chromosome number is 6 in Culex pipiens and both \(D\) and \(M\) loci are linked on the same chromosome. (b) How might a sex-ratio distorter such as \(d d\) be used to control pest population numbers?

What type of evidence supports the conclusion that the primary sex ratio in humans is much higher than the secondary sex ratio?

In chickens, a key gene involved in sex determination has recently been identified. Called \(D M R T 1\), it is located on the \(Z\) chromosome and is absent on the W chromosome. Like \(S R Y\) in humans, it is male determining. Unlike \(S R Y\) in humans, however, female chickens (ZW) have a single copy while males (ZZ) have two copies of the gene. Nevertheless, it is transcribed only in the developing testis. Working in the laboratory of Andrew sinclair (a co- discoverer of the human \(S R Y\) gene), Craig Smith and colleagues were able to "knock down" expression of \(D M R T 1\) in \(Z Z\) embryos using RNA interference techniques (see Chapter 17 ). In such cases, the developing gonads look more like ovaries than testes [Nature 461: 267 (2009)]\(.\) What conclusions can you draw about the role that the \(D M R T 1\) gene plays in chickens in contrast to the role the SRY gene plays in humans?

Describe how nondisjunction in human female gametes can give rise to Klinefelter and Turner syndrome offspring following fertilization by a normal male gamete.
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