39 A. When a female tortoise shell cat mates with a black male cat, the resulting offspring that could be produced have the following genotypic frequency: XBXb: XBY:XbXb:XbY with a 1:1:1:1 ratio.
39 B. This is the result of mosaic expression. Female tortoise shell cats have a mixed coat color in heterozygous individuals. This due to the expression of both the dominant allele (XB) resulting in patches of their coat being orange and the recessive allele (Xb) which results in patches of their coat being black. This is a result of one of the X chromosomes being inactivated during development. So some epithelial cells will have one gene inactivated and others will have the other gene inactivated. In the inheritance for the next generation, the X chromosome donated by the mother cat could therefore be either the dominant or recessive allele for the coat color gene.
40 A. If a RNA transcript from a protein coding gene contains 1200 nucleotide bases, it will contain 400 codons including the stop and start codons. A codon is comprised of three nucleotide bases that code for different amino acids or start/stop messages during translation. So therefore, therefore if the transcript contains 1200 bases, you simply divide by three to get the number of codons.
40 B. The resulting protein should be 398 amino acids. There needs to be both a start and stop codon to tell when the transcript should start and stop, so we need to deduct these from the total of 400.
40 C. When a gene only produces a protein that is 100 amino acids long in muscle cells versus producing a protein that has 225 amino acids in the lining of the intestine, this is an example of alternative splicing. Multiple proteins can be produced by the same gene due to alternative splicing. During transcription, an exon can either be included or not included in the messenger RNA. This will result in the code being translated differently, resulting in a different number of amino acids, therefore a different protein.
41 A. Since the heterozygotes have the dominant trait, this pedigree is an example of a dominant trait.
41 B. Genotypes of the individuals in the pedigree:
- Pp (heterozygous)
- Pp (heterozygous)
- pp (homozygous recessive)
- PP (homozygous dominant) or Pp (heterozygous)
- pp (homozygous recessive)
- pp (homozygous recessive)
- PP (homozygous dominant) or Pp (heterozygous)
- Pp (heterozygous)
- Pp (heterozygous)
- Could be PP (homozygous dominant), Pp (heterozygous), or pp (homozygous recessive)
41 C. If individual 7 and 8 are planning to have a child, there are two possibilities. The mother could either be homozygous dominant or heterozygous. The father is heterozygous. When doing a Punnett square analysis of both possibilities, we get the following results:
- Mother is homozygous dominant:
- Genotypic ratio is 1:1, homozygous dominant to heterozygous
- Phenotypic ratio is 100% for the dominant trait
- Mother is heterozygous:
- Genotypic ratio is 1:2:1, homozygous dominant to heterozygous to homozygous recessive
References
Alternative Splicing. (2013, Feb. 22). In Wikipedia. Retrieved 2013, Mar. 23. http://en.wikipedia.org/wiki/Alternative_splicing
The Genetics of Calico Cats. (nd.). General format. Retrieved 2013, Mar. 23. http://www.bio.miami.edu/dana/dox/calico.html
