Which kind of graph line shows logistic growth
1 answer:
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Hello. This question is incomplete. Also, you forgot to show the flowchart. The flowchart is attached below and the full question is:
The flowchart below shows the three generations of a cross between a pea plant that has yellow pods and a pea plant that has green pods. Green pods are the dominant trait. The flowchart is missing the labels that describe the traits.
In which squares should the phrase “Green pods” appear?
1.A and D 2.B and E 3.A,C and D 4.A,B,C,D and E
Answer:
3.A,C and D
Explanation:
As shown in the question above, the flowchart shows the crossing of a pea plant with dominant features (green pods - AA) and a pea plant with recessive features (yellow features - aa). The crossing between plants with AA and aa alleles generates a completely Aa population, which in this case, has the dominant characteristic, that is, it has green pods. This is because the "Aa" alleles are called heterozygous and develop the dominant characteristic.
As we can see in the flowchart, the crossing between the two pea plants generated an offspring that is identified by table C, as we know this offspring has green pods and in the flowchart it is represented by a grayish rectangle. Therefore, we can say that the other gray rectangles represent pea plants with green pods, which are rectangles A, C and D.
Complete question:
Suppose "A" is a dominant gene for the ability to taste phenylthiocarbamide and "a" is a recessive gene for the inability to taste it. Which couples could possibly have both a child who tastes it and a child who does not?
a. father AA, mother aa
b. father Aa, mother AA
c. father Aa, mother Aa
d. father AA, mother AA
Answer:
c. father Aa, mother Aa
Explanation:
According to the given information, the ability to taste phenylthiocarbamide is a dominant trait and is imparted by the allele "A". This phenotype would be expressed in both homozygous and heterozygous conditions. The non-taster phenotype would be expressed in the homozygous recessive genotypes only.
To have both taster and non-taster children, both the parents should have at least one copy of the recessive allele. Among the given options, the father with genotype Aa and the mother with genotype Aa have the possibility to have both taster and non-taster children.
Aa x Aa= 3/4 taster (1/4 AA and 1/2 Aa): 1/4 non-taster (1/4 aa)
