basic genetics

[ihatethis]

ok, so im vary frustrated with this...please help me. i already know the answer, i just dont know how to get it.

 

a trihybrid pea plant, having the genotype AaBbCc is self fertilised. all loci are unlinked. there is complete dominance at the A and B loci, but incomplete dominance at the C locus. the fraction of progeny that will be phenotypically diffrent from the parent is...

 

the correct answer is 23/32. how do i get this answer?

 

also here is another one im stuck on,

 

phenylketonuria, a metabolic didorder in humens, is inherited a san autosomal recessive trait. a husband and a wige, noth hererozygous for the gene, plan to have six childred. what is the probability that, in any order, gour of the offspring will be normal and two will have phenylketonuria?

 

please tell me the steps to solving these problems if you know them.

thanks

[jimmydasaint]

Hi,

I hope you got the answer eventually. I am not allowed to explain the answers. However, I can point you to a website and ask you to read their answer:

 

http://www.saylor.org/site/wp-content/uploads/2011/07/A-Trihybrid-Cross-Example-Using-Mendel%E2%80%99s-Sweet-Peas.pdf

 

Also, there may be some confusion about incomplete dominance (partial expression from both alleles rather than complete expression from both) from your question so I have this explanatory link:

 

http://biology.about.com/od/geneticsglossary/g/incompletedom.htm

 

Finally, phenylketonuria is autosomal recessive. This means that it is not linked to X chromosomes and is a simple monohybrid cross. The outcomes are probabilities and if the probability of a child inheriting phenylketonuria from 2 heterozygous parents is 0.25, go back to your Maths books and see how to work out the probability of independent events.

 

http://www.stsci.edu/~tbeck/genetics.html

 

I am not allowed to help more, but if this is not helpful, ask me for more vague hints...