Answer:
1/8
Explanation:
Given that the trihybrid parents have AaBbCc genotype for fruit color. The trait is a quantitative trait i.e. each dominant allele will have an additive effect on it. In this case, AaBbCc and AABBCC will not produce same fruit color because AaBbCc has only three loci contributing to the color while in AABBCC all the six loci are contributing to the color. For an offspring to be exactly similar to the AaBbCc parents it should have the same genotype of AaBbCc.
The probability of Aa to come from a cross between Aa and Aa is 2/4 or 1/2
The probability of Bb to come from a cross between Bb and Bb is 2/4 or 1/2
The probability of Cc to come from a cross between Cc and Cc is 2/4 or 1/2
So the collective probability of AaBbCc offspring from a cross between AaBbCc and AaBbCc parents would be=
1/2 * 1/2 * 1/2 = 1/8
Hence, assuming no effects of the environment, 1/8 of the offspring will have the same fruit color phenotype as the trihybrid parent.
Answer:
Evidence for evolution comes from many different areas of biology:
Anatomy. Species may share similar physical features because the feature was present in a common ancestor (homologous structures).
Molecular biology. DNA and the genetic code reflect the shared ancestry of life. DNA comparisons can show how related species are.
Biogeography. The global distribution of organisms and the unique features of island species reflect evolution and geological change.
Fossils. Fossils document the existence of now-extinct past species that are related to present-day species.
Direct observation. We can directly observe small-scale evolution in organisms with short lifecycles (e.g., pesticide-resistant insects).
Https://quizlet.com/17718770/chapter-8-aquatic-biodiversity-flash-cards/
Answer:
The correct answer will be options- A, B and D.
Explanation:
The metric system is an internationally accepted decimal system of the measurement which was adopted in France in 1795. This system is now widely accepted and used system in the world.
The metric units are based on the division and multiplication by an integer of ten thus, units can be scaled on the 10 which makes the calculation easy.
The metric system was accepted as the international system of units in 1875 when the International Bureau of Weights and Measures was established. The motto of this system was that it could be followed by all people all the time.
Thus, the selected options are the correct answers.
Answer:
The frequency of A1 be on Big Pine Key after a single generation of migration from No Name Keyp is 0.2276
Explanation:
Whenever it occurs migration between two populations, there is genetic flux going on. Genetic flux might be considered as an evolutive strength only if migration > 0 and if the allelic frequency in one generation is different from the allelic frequency in the next generation.
Genetic flux acts homogenizing the allelic frequencies between the two populations, and it might introduce variability.
By knowing the allelic frequencies in both populations at a certain time and the migration rate, we can calculate the allelic frequencies of populations in the next generation. This is:
pA₂=pA₁(1-m)+pB₁ m
pB₂=pB₁(1-m)+pA₁ m
Being
- A one population and B the other population
- pA₁ and pB₁ the frequencies of the p allele before migration,
- pA₂and pB₂ the frequencies of the p allele after migration,
- m the migration rate
In the exposed example, we know that:
- No Name Key population allelic frequency: A1 = 0.43 and A2 = 0.57
- Big Pine Key population allelic frequency: A1 =0.21 and A2 = 0.79
Let´s say that p represents A1 allele, and q represents A2 allele.
The frequency of A1 allele (p) be on Big Pine Key (Population B) after a single generation of migration from No Name Key (Population A)
pB₂=pB₁(1-m)+pA₁ m
pB₂=0.21 x (1 - 0.08) + 0.43 x 0.08
pB₂= 0.2276
The allelic frequency in a population after one generation is the allelic frequency of individuals of that population that did not migrate (21 x (1 - 0.08) plus the allelic frequency of the new individuals that came from the other population (0.43 x 0.08).
You can corroborate your result by calculating the q allele frequency in the next generation and summing both of them up. The result should be one.
qB₂= qB₁(1-m)+qA₁ m
qB₂= 0.79 x (1 - 0.08) + 0.57 x 0.08
qB₂= 0.7724
p + q = 1
0.2276 + 0.7724 = 1
