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In some populations of grasses, the ability to grow in soil contaminated w/ toxic metals like nickel is a dominant trait determined by the gene "R".
(a) In a population at genetic equilibrium, 51% of the seeds can germinate in contaminated soil. What are the frequencies of "R" and "r"?
(b) What percentage of the plants that germinate in the contaminated soil will be heterozygous?

2- Cystic fibrosis is a genetic disease of humans caused by homozygosity for a recessive gene called the gene "a". In a particular large population, what proportion of the population can be expected to be carriers of this trait (heterozygous)?
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I'm still working on these, so I haven't come up with anything yet, but I'll post what I get as I get them. Meanwhile, can someone explain these so I know if I'm on the right track? I would really appreciate it. Thank You! = )

2007-02-14 09:55:40 · 1 answers · asked by Miss*Curious 5 in Science & Mathematics Biology

Cystic fibrosis is a genetic disease of humans caused by homozygosity for a recessive gene called the gene "a". In a particular large population, the newborns affected with cystic fibrosis is 1 in 3,600. In this population what proportion of the population can be expected to be carriers of this trait (heterozygous)?

2007-02-14 10:23:09 · update #1

1 answers

You have a lot of H-W problems ...

1. a. "Can germinate" is the dominant trait, and those individuals will be RR and Rr. If R is represented in H-W by letter p ...
pp + 2pq = 0.51
That leaves qq which must be 1 - 0.51 = 0.49
qq = 0.49
q = 0.7 = frequency of r allele
p = 1 - 0.7 = 0.3 = frequency of R allele

b. heterozygous ones are 2pq = 2 * 0.3 * 0.7 = 0.42
42% are heterozygous

2. This one needs some data. Do you know the frequency of cystic fibrosis?

2007-02-14 10:08:38 · answer #1 · answered by ecolink 7 · 0 0

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