Answer: For three traits inherited in a dominant/recessive pattern, the probability of an offspring in a trihybrid cross between parents heterozygous for all three traits to be recessive at exactly two of the three loci is 23/64.
Explanation: Let A, B and C represent the dominant alleles and a, b and c represent the recessive alleles. The genotype of each of the parent will be AaBbCc since it has been said to be heterozygous for all the three traits. There are eight (8) possible different types of gametes from each of the genotype. This can be calculated using 2^n where n is the number of heterozygous loci.
The number of heterozygous loci in the genotype AaBbCc is three (3). 2^3 = 8. The gametes are;
ABC
ABc
AbC
Abc
aBC
aBc
abC and
abc.
Therefore, sixty-four possible offsprings are expected from the cross AaBbCc x AaBbCc.
Out of these offsprings, twenty-three (23) are expected to be heterozygous at two loci as boldly shown in the attached image.
Answer:
If there is homologous chromosomes (metaphase I) or duplicated chromosomes/sister chromatids (metaphase II) in the middle of the cell.
Explanation:
Meiosis involves two series of nuclear divisions grouped into meiosis I and meiosis II. Each division has the same number of stages i.e prophase, metaphase, anaphase, telophase etc. Meiosis I involves the separation of homologous chromosomes i.e similar but non-identical chromosomes from each parent.
On the other hand, meiosis II involves the separation of sister chromatids (duplicated chromosome). Since METAPHASE is generally characterized by the alignment of chromosome at the middle of the cell for separation in the anaphase stage, it means that the difference between metaphase in meiosis I and II will be whether it is homologous chromosomes that are in the middle or sister chromatids.
Therefore, according to this question, I would know if the cartoon is in metaphase I or II if:
- there are homologous chromosomes in the middle of the cell (metaphase I)
- there are sister chromatids in the middle of the cell (metaphase II).
Answer:
B) a nonsense mutation; this is because a nonsense mutation results in the change of a regular amino acid codon into a stop codon, which ceases translation. This fits with the problem's description of the protein that causes the symptoms as too short, as translation is the process by which proteins/polypeptides are created. A missense mutation would not be the answer because it still codes for an amino acid, which would not shorten the protein. A duplication of the gene would probably just lengthen the protein or not affect its length at all.
If from the low tide to the high tide there was a break of 6h and 4m than the next tides will happen at the same rate:
8:35 + 6:04 = 14:39 = next low tide.
14:39 + 6:04 = 20:43 = next high tide.
Note that two same tides occur about 12h to 12h and 25m apart, and that is the case here:
14:39 - 2:31 = 12:08 hours apart
20:42 - 8:35 = 12:07 hours apart.
Hope it helped,
Happy homework/ study/ exam!
