This is the day to day problem of a medical microbiology laboratory: the "tube containing a liquid nutrient medium" in the real world could, for example, represent blood culture bottles (aerobic or anaerobic). The basic process is as follows: a loop is first flamed to ensure sterility and then used to transfer a small portion of bacterial-laden liquid media to the agar medium by streaking it across the surface of a sterile petri dish. The dish is then covered, inverted, and placed in an appropriate incubator. When colonies begin to form on the surface of the agar plate, individual colonies can be taken up using the sterile loop for analysis or further propagation. The assumption is that a single colony represents growth from a single bacterium.
Answer:
In Amoeba and Paramoecium, osmoregulation occurs through contractile vacoule. Osmoregulation is a phenomenon in which contractile vacuole plays an important role in maintaining the water balance of the cell. Paramoecium contains two contractile vacuoles which have fixed position.
Answer:
If KE = 0, the pendulum is not moving. At the top of the swing, the pendulum must stop for a brief moment.
Explanation:
The common or popular term used to define an 'anticipated length of a person's life' is Life Expectancy. However other terms can be used such as Life Span, Longevity or a person's period of existence.
Regardless of the term used, the calculation of a person's anticipated length of life should remain the same. This should include: birth year, current age, and use of demographic factors including gender, sex, race etc.
Answer:
0.9342
Explanation:
The Hardy-Weinberg equation states that p² + 2pq + q² = 1,
where p is the frequency of the dominant 'normal' (n) allele and q is the frequency of the recessive 'albino' (a) allele in the population, while q² represents the frequency of the homo-zygous albino genotype (aa), p² represents the frequency of the homo-zygous normal genotype (nn) and 2pq represents the frequency of the heterozygous genotype (na).
In this case, the frequency of individuals in the population that have the genotype aa (q²) is equal to 26/6000 = 0.004333. In consequence, q is equal to √ 0.004333 = 0.0658.
Moreover, the allele frequency of the normal (n) allele p is equal to 1 - q = 1 - 0.0658 = 0.9342, so p² (nn) = (0.9342)² = 0.8727.
Finally, the frequency of the heterozygous genotype (na) is 2pq = 2 x 0.9342 x 0.0658 = 0.123.
