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9 years ago
In a species of fern, there is a trait called branched dimorphism, in which the branches show two different patterns, as opposed to wildtype fern that only show one branching pattern. Two true-breeding parents, one single branching and the other with the dimorphic pattern, were crossed. All the F1 offspring displayed the single branching pattern. When the F1 plants were crossed, the F2 offspring were counted and it was found that there were 112 single branching offspring and 8 dimorphic branching offspring. How can you explain these results?
Please show all work and why this odd ratio occurs!!
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9 years ago
First of all, since all F1 offsprings displayed single branching pattern, the allele for single pattern should be the dominant. (That's because the alleles that passed from the dimorphic branched parent were not expressed to the offsprings). Let's name the alleles S the dominant single branch and s the recessive dimorphic branch.

Since the ratio is odd, we should thing a bit differently.

My first though is, maybe there are more than one gene responsible for the trait (most likely 2 genes/4 alleles)

So, let's assume that there are 2 genes (4 alleles) responsible for the branch trait. We also must assume that the genes are independent which means they are inherited separately and not together.

We will rename the alleles: S and s for the 1st gene, and S2 and s2 for the 2nd gene. We know that S and S2 are dominant, so even one of them (any of them) will make the plant single branched. Now we will make the crossings, just like any crossing with 2 independent genes:

In the first crossing we have:
SS S2S2 x ss s2s2
offsprings: 100% Ss S2s2 all are single branched

Next crossing
F1  Ss S2s2 x Ss S2s2

               SS2      Ss2      sS2      ss2
SS2    SSS2S2  SSS2s2  SsS2S2  SsS2s2
Ss2    SSS2s2   SSs2s2  SsS2s2  Sss2s2
sS2    SsS2S2   SsS2s2  ssS2S2  ssS2s2
ss2    SsS2s2    Sss2s2  ssS2s2  sss2s2

The only case where there are no dominant allele is the genotype sss2s2 which is 1/16 possibility of dimorphic pattern plants. All other genotypes give single branched plants.
As you see, this possibility matches with the 8/(112+8) = 8/120 = 1/15 which is pretty close to the 1/16.

Also next time use a more comprehensive title.
wrote...
9 years ago
An artery that has been exposed to a lifetime of unhealthy food is slowly building up atherosclerotic plaque. If the vessel diameter is 0.8 mm, and the blood flow rate through the vessel is 7 mL/sec, and the plaque is building up uniformly on the wall of the vessel,
a) Calculate how thick the plaque needs to be change the velocity of blood flowing through the vessel by 30%. Would the velocity decrease or increase?
b) Plot the velocity of blood through the vessel and the plaque thickens from 0 to Tx10^-11 cm where T is 912147102. Please provide raw data along with the plot.
nanduanandu,  Puja Ghosh,  prakash,  mkgoel20
wrote...
9 years ago
You are seeing a new patient: A 42 year old male with no major health issues to date has seen a 17 pound gain in weight over the last month. He has been eating from a “fast-food” restaurant regularly and is feeling lethargic. Physical examination indicates muscle weakness. Laboratory results indicate an enlarged liver with excessive glycogen granules and high levels of fat (abnormal fat metabolism), negative for hepatitis and low testosterone levels. He also has abnormal calcium balance in his cells.
What specific organelle is malfunctioning for this patient? Discuss your rationale for your answer (this should be precise and ordered logically). Also discuss what you would advise this patient to do. Although the fast food is an obvious reason for gaining weight, there is an organelle that should still help in metabolizing any food.
wrote...
9 years ago
Part A - Recombination in dihybrid crosses
You have screened for several new recessive mutations in a species of wasp.
•   Wasps homozygous for apricot (aa) have pale orange eyes. (Wild-type eyes are brown.)
•   Wasps homozygous for blunt (bb) have short wings. (Wild-type wings are long.)
You make a pure-breeding double-mutant (apricot, blunt) line and cross it with wild-type wasps. The F1is wild-type in appearance.
You testcross the F1 dihybrids with the double-mutant line (the “tester” genotype) and obtain four phenotypes in the F2:
•   wild type
•   apricot, blunt
•   apricot
•   blunt
Determine the following for each F2phenotype:
•   the haploid genotype of the gamete it received from the F1 dihybrid
•   its full diploid genotype
•   whether the gamete it received from the F1dihybrid was recombinant (REC) or nonrecombinant (nonREC).
  
anithaisro,  prakash,  gk.forest,  styler1979
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