Answer:
7%.
Explanation:
The recombinant progeny might occur due to the exchange of genetic material between non sister chromatids known as crossing over during the meiosis.
The chromosome sequence is DABC. A and B distance is 4 map units, B and C is 2 map units, B and D is 5 map units. The distance between A and D is 1 map unit that can be calculated by subtracting the distance between BD and AB ( 5-4). The CD map distance can be calculated by adding the distance between AD, AB and BC ( 1 + 4+ 2). The recombination frequency is equal to distance between them is 7%.
Thus, the answer is 7%.
The correct answer is emphysema.
Emphysema is the enlargement of air sacs in the lungs.
It is a long-term, progressive disease of the lungs that primarily causes shortness of breath. People who have emphysema, their air sacs in the lungs (alveoli) is damaged. Over a period of time, the inner walls of the air sacs weaken and rupture creating larger spaces instead of many small ones. Normally, the lung tissue holds these small airways called bronchioles, open, allowing air to leave the lungs on exhalation. However, when the lung tissue is damaged like in the case of emphysema, these airway collapse, making it difficult for the lungs to empty and the air (gases) becomes trapped in the alveoli thus causing impaired gas exchange.
Function. There, the villi and the microvilli increase intestinal absorptive surface area approximately 30-fold and 600-fold, respectively, providing exceptionally efficient absorption ofnutrients in the lumen. There are alsoenzymes (enterocyte digestive enzyme) on the surface for digestion.
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