Answer : Without its keystone species, the ecosystem would be dramatically different or cease to exist altogether.
Answer:
a. Ed/eD
b. RF=0.12
Explanation:
The alleles for genes D/d and E/e are:
- D_: only on neck spikes
- dd: along back spikes
- E_: long spikes
- ee: short spikes
After testcrossing a double heterozygote (DdEe x ddee) there are 4 types of offspring, two of them much more abundant than the other two. The homozygous recessive parent can only produce <em>ed</em> gametes, so the phenotypes of the offspring depend on the gametes that the double heterozygous parent produced.
The offspring was:
- Ed/ed 79
- ed/ed 12
- ED/ed 10
- eD/ed 76
Total: 177
a) This result suggests that the genes are linked. Since recombination is a rare event, the most abundant phenotypes always come from the parental gametes, and the least abundant come from the recombinant gametes.
<u>Therefore, the genotype of the doube heterozygote parent was </u><u><em>Ed/eD</em></u><u>.</u>
b) Recombination frequency (RF) = Recombinants / Total
RF = (12+10)/177
<u>RF = 0.12</u>
Answer:
An individual whose genotype is heterozygous cannot be for the recessive
Explanation:There are two possible genotypes either homozygous dominant or heterozygous dominant. For a charecter there can be two gene .One is recessive and other one is dominant. In presence of the dominant gene the recessive gene of the same character cannot express . IF recessive gene is homozygous then only it will express .Therefore for the individual it is heterozygous for the dominant
There are four bases found in DNA: adenine (A), cytosine (C), guanine (G), and thymine (T). Adenine forms a base pair with thymine, and cytosine forms a base pair with guanine. There is a one-to-one relationship in these base pairings (Chargaff’s rule), which means that if you know the percentage of any one of them within a given DNA sample, you can calculate the percentages of the other three. In this case, you're given the percentage of guanine, and you want to find out the percentage of adenine.
Since guanine base-pairs with cytosine and since there must be as much cytosine as there is guanine, 41% of the bases in this gene are cytosine as well. That means that adenine and thymine <em>together </em>make up the remaining 18% (100% − 41% G − 41% C) of the base pairs. If there must be an equivalence in the number of thymine and adenine bases per Chargaff's rule, then half of the remaining base pairs must comprise adenine and the other half comprise thymine. Half of 18% is 9%.
Thus, adenine makes up 9% of the bases in this gene.
Answer:
d) stomata
Explanation:
CAM plants lose their stomata during the day and open them at night.
Therefore, CAM plants do not lose thylakoids, chloroplasts, or vascular bundles during the day and open them at night.
