How could you do the following questions without drawing thepunnett squares? 1.

Question

How could you do the following questions without drawing thepunnett squares? 1. If parents are heterozygous for all three traits, what isthe probability that an offspring is heterozygous for all threetraits? a.) 1/64 b.) 2/64 c.) 4/64 d.) 8/64 e.) 32/64 2. If the parents were heterozygous for all three traits, whatis the probability that an offspring is heterozygous for one traitand homozygous recessive for one trait, and homozygous dominant forone trait? a.) 1/64 b.) 2/64 c.) 4/64 d.) 8/64 e.) 32/64 How could you do the following questions without drawing thepunnett squares? 1. If parents are heterozygous for all three traits, what isthe probability that an offspring is heterozygous for all threetraits? a.) 1/64 b.) 2/64 c.) 4/64 d.) 8/64 e.) 32/64 2. If the parents were heterozygous for all three traits, whatis the probability that an offspring is heterozygous for one traitand homozygous recessive for one trait, and homozygous dominant forone trait? a.) 1/64 b.) 2/64 c.) 4/64 d.) 8/64 e.) 32/64

Explanation / Answer

To solve genetics problems without a punnet square, you will needto use statistics. 1. Okay. You can start by making a list to keep things straight.The numbers are the probability of that particular allelecombination. 0.25=25%=1/4, each has a 1 in 4 possibility. AA 0.25      BB   0.25       CC   0.25 aa    0.25     bb     0.25      cc     0.25 Aa   0.50     Bb    0.50      Cc    0.50 The heterozygous alleles have a 0.50 probability because they canbe either Aa or aA. So 0.25+0.25=0.50. Each pair of letters is a gamete combination, so all that is needednow is to multiply the heterozygous probabilities: (0.5)(0.5)(0.5)=0.125=12.5%=1/8=8/64 (the answer is d) 2. I will use the same chart from above. We need an offspring thatis heterozygous for one trait and homozygous recessive for onetrait, and homozygous dominant for one trait. The heterozygousallele combination is a big letter and a little letter, homozygousrecessive is two little letters, and homozygous dominant is two bigletters. Aa*bb*CC=(0.50)(0.25)(0.25)=1/32=2/64 (the answer is b)

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