D. Metaphase
Always remember M in Metaphase stands for Middle, Middle on the cell!
Answer:
Recessive phenotype plant
Explanation:
If the plant has recessive genotype then it will be easy to identify because it will have the recessive phenotype. However if the plant has dominant phenotype it can either have homozygous genotype or heterozygous genotype. To confirm if the plant is homozygous or heterozygous, a test cross can be done with plant having recessive phenotype.
If our test plant is homozygous dominant, it will pass one dominant allele to the next generation and all the offspring would have dominant phenotype. If our test plant is heterozygous dominant it will pass one dominant allele to half of the offspring and one recessive allele to another half so 50% of next generation will have dominant phenotype and other 50% will have recessive phenotype.
Hence by test cross (cross with recessive phenotype plant) it is possible to determine the genotype of the uncharacterized pea plant.
In scientific reasoning, a hypothesis is made before any applicable analysis has been done. A theory, on the opposite hand, is supported by evidence: it is a principle shaped as an effort to clarify things that have already been supported by knowledge<span>.
</span>For example: “It's bright outside.”
Hypothesis: A projected clarification for a development created as a place to begin for additional investigation.
Theory: A well-substantiated rationalization nonheritable through the methodology and repeatedly tested and confirmed through observation and experimentation
Full question found from other source
The F2 generation phenotypes for each cross are shown in Table 1. (See attachment) Which of the following is the mean number per cross of F2 generation offspring that are the result of crossing over?
Answer:
B, 2.2
Explanation:
The parental genotypes are long and black vs short and white. Therefore the phenotypes that result from crossing over are long and white, and short and black. (middle two rows of the table). If we add up the total number of offspring with these genotypes we get 6 long whites, and 5 short blacks.
The total is 11 from 5 crosses, so the mean is 11/5 = 2.2
The answer should be competitive inhibitors
