ACTIVITY 3, THE HARDY WEINBERG EQUATION Faoqancy of banocygn dominant indinideak

Question

ACTIVITY 3, THE HARDY WEINBERG EQUATION Faoqancy of banocygn dominant indinideak 2. Ahnikn i a neoouive gnetk sonder vhkch nwuis la the lack af dhe production of pigsest is Caklax the foquy aflected indnidialh. In dhe Unirnd Setcs. I in 17.000 popke are lbic af carricn Ovescroergotesl of ahinim. Shmw your cakulaion Challenge Question Lask sp infomation on che ngle astice Xeither ia your tcebook oe the isternet. Calco cas are mealy whire with ceange and black paschea. The eange and black colocasion is auc to the single active X Femake cats betemergous fie this trait arc calica .If a clico cat had 12k byaHask mude cat, hew mary of the kines oald be opeund to be calloa! CnstractPun sqeare Horw many kittens would be epacted io be black Esplain why cao cs are alwaps fanales.

Explanation / Answer

Answer:

1).

Number of huntington's disease bearing people = 100000 - 9339 = 661

The frequency of huntington's disease =q^2= 661/10000 = 0.0661

The frequency of diseased allele = q = 0.2571

As p+q = 1

Normal allele frequency = p = 1-q = 1-0.2571 = 0.7429

Frequency of heterozygotes = 2pq = 2 * 0.7429 * 0.2571 * 10000 = 3820

Fequency of homozygous dominant individuals = p^2 = 0.7429 * 0.7429 * 10000 = 5519

Frequency of homozygous recessive individuals = 661

2).

Frequency albino = q^2 = 1/17000 = 0.00006

q = 0.008

p+q = 1

p = 1-q = 1-0.008 = 0.992

The frequency of carriers (heterozygotes) = 2pq = 2 * 0.992 * 0.008 = 0.01587

The carrier individuals = 0.01587 * 17000 = 270

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