A client was diagnosed with pernicious anemia due to poor food intakes. What food groups is she likely lacking?
1 answer:
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Answer:
The most likely genotype of the male parent is RrBbmm.
Explanation:
First of all, we have 3 genes and 6 alleles:
For color pattern:
- RR or Rr: Solid tail color
- rr: Spotted tail
For coat color:
- BB or Bb: Black coat color
- bb: Brown color
Expression of gene B/b depends on M:
- MM or Mm: B/b is expresed.
- mm: B/b is not expresed, yellow coat.
So, a cross between a solid yellow-tailed male and a solid brown-tailed female was made, and the offspring was:
- 6 solid yellow
- 2 spotted yellow
- 3 solid black
- 1 spotted black
- 3 solid brown
- 1 spotted brown
Total: 16
If it was produced all possible phenotypes, it means that parents have to be carriers of R/r, B/b, and M/m alleles. Male parent is yellow, so he has to be mm for this gene, but for the two other alleles he should be heterozygous:
RrBbmm
And female parent is solid brown-tailed, so she has M allele, (probably heterozygous), and Rr and bb (color brown),, like this:
RrbbMm
When these two genotypes are crossed, 8 gametes are generated for each parent, and 64 possible combinations. This cross in the attached Punnett Square.
As a result, the phenotypicial ratios are in the attached table and they are the same as the reported in the question:
Punnett Square
- 24 (6) solid yellow
- 8 (2) spotted yellow
- 12 (3) solid black
- 4 (1) spotted black
- 12 (3) solid brown
- 4 (1) spotted brown
Total: 64
