Given a context passage, generate a question from the passage such that its answer is shortest continous span from the passage.

Ex Input:
Facioscapulohumeral muscular dystrophy (FSHD) is a dominant disease linked to contraction of an array of tandem 3.3-kb repeats (D4Z4) at 4q35. Within each repeat unit is a gene, DUX4, that can encode a protein containing two homeodomains. A DUX4 transcript derived from the last repeat unit in a contracted array is associated with pathogenesis but it is unclear how. Using exon-based microarrays, the expression profiles of myogenic precursor cells were determined. Both undifferentiated myoblasts and myoblasts differentiated to myotubes derived from FSHD patients and controls were studied after immunocytochemical verification of the quality of the cultures. To further our understanding of FSHD and normal myogenesis, the expression profiles obtained were compared to those of 19 non-muscle cell types analyzed by identical methods. Many of the ~17,000 examined genes were differentially expressed (>2-fold, p<0.01) in control myoblasts or myotubes vs. non-muscle cells (2185 and 3006, respectively) or in FSHD vs. control myoblasts or myotubes (295 and 797, respectively). Surprisingly, despite the morphologically normal differentiation of FSHD myoblasts to myotubes, most of the disease-related dysregulation was seen as dampening of normal myogenesis-specific expression changes, including in genes for muscle structure, mitochondrial function, stress responses, and signal transduction. Other classes of genes, including those encoding extracellular matrix or pro-inflammatory proteins, were upregulated in FSHD myogenic cells independent of an inverse myogenesis association. Importantly, the disease-linked DUX4 RNA isoform was detected by RT-PCR in FSHD myoblast and myotube preparations only at extremely low levels. Unique insights into myogenesis-specific gene expression were also obtained. For example, all four Argonaute genes involved in RNA-silencing were significantly upregulated during normal (but not FSHD) myogenesis relative to non-muscle cell types. DUX4's pathogenic effect in FSHD may occur transiently at or before the stage of myoblast formation to establish a cascade of gene dysregulation. This contrasts with the current emphasis on toxic effects of experimentally upregulated DUX4 expression at the myoblast or myotube stages. Our model could explain why DUX4's inappropriate expression was barely detectable in myoblasts and myotubes but nonetheless linked to FSHD.

Ex Output:
Which disease is associated with the ectopic expression of the protein encoded by the gene DUX4?


Ex Input:
Facioscapulohumeral muscular dystrophy (FSHD) is a primary muscle disorder with autosomal dominant inheritance. FSHD was mapped to chromosome 4 locus q35, but the gene is not yet known. It is characterised by progressive, often asymmetric, selective muscular weakness and great clinical variability. The aim of the study was to analyze 62 FSHD cases from 44 Polish families in which the diagnosis was confirmed by DNA analyses. FSHD diagnosis was based on the clinical findings and standardized investigations confirming primary muscular involvement (EMG, muscle biopsy). DNA analysis was based on EcoRI/BlnI restriction enzyme digestion followed by hybridization with P13E-11 molecular probe. In our material, we have found a relatively large percentage (41%) of big deletions (EcoRI/BlnI fragment of 10-15 kb [kilo bases]), which in the majority of cases (67%) was present in isolated cases. In 10 families (23%) the phenotype was assessed as severe. These are cases with the onset before the age of 10 and fast progression. "Middle sized" deletions (EcoRI/BlnI fragment of 16-29 kb) were prevalent in familial cases and present in 57% of families. "Small" deletion was found in one family (EcoRI/BlnI fragment of 30 kb). Somatic mosaicism was confirmed in one case. De novo mutations were shown in 11% of the examined families. The results of this study indicate that the bigger the deletion, the more severe the FSHD course, however there are some exceptions. A similar relationship has been shown by previous research. Molecular analyses are particularly important in atypical and sporadic cases. It is the first genetic presentation of a group of patients' with this kind of dystrophy in the Polish population.

Ex Output:
What is the mode of inheritance of Facioscapulohumeral muscular dystrophy (FSHD)?


Ex Input:
What determines if a baby is a male or female? Recall that you have 23 pairs of chromosomesand one of those pairs is the sex chromosomes. Everyone has two sex chromosomes. Your sex chromosomes can be X or Y. Females have two X chromosomes (XX), while males have one X chromosome and one Y chromosome (XY). If a baby inherits an X chromosome from the father and an X chromosome from the mother, what will be the childs sex? The baby will have two X chromosomes, so it will be female. If the fathers sperm carries the Y chromosome, the child will be male. Notice that a mother can only pass on an X chromosome, so the sex of the baby is determined by the father. The father has a 50 percent chance of passing on the Y or X chromosome, so there is a 50 percent chance that a child will be male, and there is a 50 percent chance a child will be female. This 50:50 chance occurs for each baby. A couples first five children could all be boys. The sixth child still has a 50:50 chance of being a girl. One special pattern of inheritance that doesnt fit Mendels rules is sex-linked inheritance, referring to the inher- itance of traits that are located on genes on the sex chromosomes. Since males and females do not have the same sex chromosomes, there will be differences between the sexes in how these sex-linked traitstraits linked to genes located on the sex chromosomesare expressed. Sex-linked traits usually refer to traits due to genes on the X chromosome. One example of a sex-linked trait is red-green colorblindness. People with this type of colorblindness cannot tell the difference between red and green. They often see these colors as shades of brown ( Figure 1.1). Boys are much more likely to be colorblind than girls ( Table 1.1). This is because colorblindness is a sex-linked, recessive trait. Boys only have one X chromosome, so if that chromosome carries the gene for colorblindness, they will be colorblind. As girls have two X chromosomes, a girl can have one X chromosome with the colorblind gene and one X chromosome with a normal gene for color vision. Since colorblindness is recessive, the dominant normal gene will mask the recessive colorblind gene. Females with one colorblindness allele and one normal allele are referred to as carriers. They carry the allele but do not express it. How would a female become colorblind? She would have to inherit two genes for colorblindness, which is very unlikely. Many sex-linked traits are inherited in a recessive manner. Xc Xc X (carrier female) Xc Y (colorblind male) X Y X XX (normal female) XY (normal male) According to this Punnett square ( Table 1.1), the son of a woman who carries the colorblindness trait and a male with normal vision has a 50% chance of being colorblind.

Ex Output:

most sex-linked genes are ___________.