Individual plants homozygous for various combinations of linked R and T (i.e., RRTT, RRtt, rrTT, rrtt) were cross in pairs to produce F1 plants, which were then self-fertilized to produce an F2 generation. R is dominant to r and T is dominant to t. State the genotype (e.g., RT/rt) for each F1 parent of the following F2 phenotypic distribution: F2 Phenotype distribution F1 Parent Genotype RT Rt rT rt 106 34 0 0 ________________ 0 0 104 36 ________________ 99 0 31 0 ________________ 64 10 8 18 ________________ Solution It can be seen from the quantitative data that the number of offsprings for various combinations has been given into 4 sets. Each of these sets represent a population of offsprings with their respective genotypic ratios. According to the information, it is also clear that the individual plants or P1 population was homozygous for both the alleles, i.e. the genotypes of parents were RRTT and rrtt. Thus, after crossing these parents, the F1 offsprings should all have RrTt genotype. However, since the R and T genes are linked, the ratio of this cross will deviate from normal Mendelian poulations. Thus, the answers and explanations can be found as below: _____________________________________________________________________________ __________________ F2 Phenotype distribution F1 Parent Genotype Explanation _____________________________________________________________________________ __________________ RT Rt rT rt RRTt * RrTt Since majority of the populatio contains RT or Rt 106 34 0 0 genotypes, it is clear that both the parents contain atleast one R allele. Since R and T alleles are linked, it suggests that at least one parent contains a dominant allele pair for both alleles. Finally, since there is no population with recessive genotypes, none of the parents can have a homozygous recessive pair of alleles. 0 0 104 36 rrTT * rrTt Since none of the offsprings carry dominant R allele, it is clearly absent in the parents as well. Since te majority of population carries dominant T allele, it must be atleast heterozygous in nature. Rest of the pouplation will carry homozygous dominant T allele. 99 0 31 0 RRTT * RrTT Since there is no offspring with recessive t allele, both parents should be homozygous dominant for this allele. Finally, since only a few offsprings carry r allele, atleast on of the parents should be heterozygous for Rr and other should be homozygous dominant. 64 10 8 18 RRTt * RrTt It can be seen that the population with recessive t allele are very few and those with dominat R allele are maximum. Thus, both the parents should contain atleast one R allele and atleast one should be homozygous. Similarly, a very few population contains t allele thus both parents should contain atleast one t allele..

1. Individual plants homozygous for various combinations of linked R and T (i.e.,
RRTT, RRtt, rrTT, rrtt) were cross in pairs to produce F1 plants, which were then self-fertilized
to produce an F2 generation. R is dominant to r and T is dominant to t. State the genotype (e.g.,
RT/rt) for each F1 parent of the following F2 phenotypic distribution:
F2 Phenotype distribution F1 Parent Genotype
RT Rt rT rt
106 34 0 0 ________________
0 0 104 36 ________________
99 0 31 0 ________________
64 10 8 18 ________________
Solution
It can be seen from the quantitative data that the number of offsprings for various combinations
has been given into 4 sets. Each of these sets represent a population of offsprings with their
respective genotypic ratios. According to the information, it is also clear that the individual
plants or P1 population was homozygous for both the alleles, i.e. the genotypes of parents were
RRTT and rrtt. Thus, after crossing these parents, the F1 offsprings should all have RrTt
genotype. However, since the R and T genes are linked, the ratio of this cross will deviate from
normal Mendelian poulations.
Thus, the answers and explanations can be found as below:
_____________________________________________________________________________
__________________
F2 Phenotype distribution F1 Parent Genotype Explanation
_____________________________________________________________________________
__________________
RT Rt rT rt RRTt * RrTt Since majority of the populatio contains RT or Rt 106 34
0 0 genotypes, it is clear that both the parents contain atleast one R allele. Since R
and T alleles are linked, it suggests that at least one parent contains a dominant allele pair for
both alleles. Finally, since there is no population with recessive genotypes, none of the parents
can have a homozygous recessive pair of alleles.

2. 0 0 104 36 rrTT * rrTt Since none of the offsprings carry dominant R allele, it is
clearly absent in the parents as well. Since te majority of population carries dominant T allele, it
must be atleast heterozygous in nature. Rest of the pouplation will carry homozygous dominant
T allele.
99 0 31 0 RRTT * RrTT Since there is no offspring with recessive t allele, both
parents should be homozygous dominant for this allele. Finally, since only a few offsprings carry
r allele, atleast on of the parents should be heterozygous for Rr and other should be homozygous
dominant.
64 10 8 18 RRTt * RrTt It can be seen that the population with recessive t allele are
very few and those with dominat R allele are maximum. Thus, both the parents should contain
atleast one R allele and atleast one should be homozygous. Similarly, a very few population
contains t allele thus both parents should contain atleast one t allele.