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1 | |
The banding patterns observed in the ultra-bar and bar-revertant flies can be explained by |
| | A) | A misaligned crossover between two X chromosomes |
| | B) | A new mutation creating a bar allele |
| | C) | The loss of an X chromosome in a female fly |
| | D) | An abnormality in chromosome staining |
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2 | |
Let's assume that a crossover requires an alignment between a minimum of one eye facet gene as shown in the figure that follows this experiment. Based on this assumption, if Bridges had begun with a true-breeding strain of ultra-bar flies, it would have been possible to obtain flies with the following number of eye-facet genes on their X chromosomes. |
| | A) | One, two, three, four, five, and six |
| | B) | Zero, one, two, three, four, and five |
| | C) | One, two, three, four, and five |
| | D) | Zero, one, two, three, four, five, and six |
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3 | |
With regard to this experiment, a "position effect" is observed because |
| | A) | A bar heterozygote has more facets than a bar homozygote. |
| | B) | A bar homozygote has more facets than an ultra-bar heterozygote. |
| | C) | A bar homozygote has more facets than an ultra-bar homozygote. |
| | D) | An ultra-bar heterozygote has more facets than an ultra-bar homozygote. |
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4 | |
The advantage of examining polytene chromosomes is |
| | A) | That's where the facet gene is found. |
| | B) | Their large size makes it possible to detect single gene duplications. |
| | C) | They are easier to stain compared to nonpolytene chromosomes. |
| | D) | They are very small, making them easy to manipulate. |
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5 | |
The data indicate that |
| | A) | Bar-eyed flies have three copies of the eye-facet gene on their X chromosome. |
| | B) | Ultra-bar eyed flies have three copies of the eye-facet gene on their X chromosome. |
| | C) | Bar-revertants have two copies of the eye-facet gene on their X chromosome. |
| | D) | Ultra-bar eyed flies have two copies of the eye-facet gene on their X chromosome. |
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