Please answer the question with x’s placed on them in its entirety . Please answ
ID: 195888 • Letter: P
Question
Please answer the question with x’s placed on them in its entirety .Please answer in its entirety
Please answer in its entirety H. Consider the following cross: Aa BB Cc Dd EE (male) X Aa Bb Cc Dd EE (female) How many cells (boxes) would there be on a Punnett square for this cross if you were to make one? 2) What t traction of the offspring from this cross would you expect to have the same genotype as the male parent? i (of the Offspirgs cot what fraction of the offspring from this cross would you expe ct to be homozygous for all five genes? Go What fraction of the offspring from this cross would you expect to be homozygous dominant for all tive ) What fraction of the offspring from this cross would yóu expect to display the dominant phenotype for 2-1% of the QAsprins coil shao donnant ngs nly 6.beno daminan five genes?
Explanation / Answer
H..1. ans= 1024 boxes in a Punnett square
Assuming that all traits exhibit independent assortment, the number of allele combinations an individual can produce is two raised to the power of the number of traits. For two traits, an individual can produce 4 allele combinations (2^2). Five traits produce 32 combinations (2^3). So in a cross each individual will have 32. So 32 x32= 1024 boxes.
H 3. Fraction of offspring homozygous for all genes
Aa x Aa= AA, Aa, aA, aa (Homozygous are AA and aa) so 2/4 or 1/2
BB x Bb= BB, Bb, BB,, Bb (Homozygous are BB which is 2/4 or 1/2
Cc x Cc= CC, Cc,, cC, cc = 2/4 or1/2
Dd xDd= DD, Dd, dD, dd= 2/4 or 1/2
EE xEE= EE,EE EE, EE= 4/4= 1
So fraction homozygous are 1/2 x1/ 2x 1/2 x1/2 x 1= 1/16
H 5. Dominant phenotype for all genes
Aa x Aa= AA, Aa, aA, aa (Dominant phenotypes will be with AA Aa and aA) so 3/4.
BB x Bb= BB, Bb, BB,, Bb Dominant phenotypes will be with BB and Bb) so 4/4=1
Cc x Cc= CC, Cc,, cC, cc =Dominant phenotypes will be with CC Cc and cC, so 3/4
Dd xDd= DD, Dd, dD, dd= Dominant phenotypes will be with DD,Dd and dD, so 3/4
EE xEE= EE,EE EE, EE= 4/4= 1
Fraction of offspring with dominant phenotypes= 3/4 x 1 x3/4 x 3/4 x 1= 27/64.
3 Albinism
Cc x Cc= CC, Cc cC and cc. So 1/4 will be albino.
a. All four are normal= 3/4 x 3/4 x 3/4 x3/4= 81/256
b. Three normal and 1 albino= 3/4 x3/4 x3/4 x 1/4= 27/256
c. Two normal and two albino= 3/4 x3/4 x 1/4 x 1/4= 9/256
d. One normal and 3 albino= 3/4 x 1/4 x1/4 x1/4= 3/256.
e. All four albino= 1/4 x1/4 x1/4 x1/4= 1/256.
Q. Probability to solve problems in genetics
AaBb x AaBb
c. Phenotype A- bb= 3/4 x 1/4= 3/16 or Genotype AAbb= 1/4 x 1/4= 1/16
So probability (A-bb or AAbb)= 3/16 + 1/16= 4/16= 1/4
d. Phenotype A-B-= 3/4 x 3/4= 9/16 or genotype aabb= 1/4 x 1/4= 1/16
So probability (A-B- or aabb)= 9/16 + 1/16= 10/16= 5/8.