Biology Homework Assignment: Analyzing Punnett Square Problems

This document presents solutions to various Punnett Square problems, a fundamental concept in genetics. The solutions cover a range of scenarios, including crosses involving homozygous and heterozygous genotypes, dominant and recessive alleles, and the prediction of offspring phenotypes. The problems address inheritance patterns for traits such as hair length, flower color, stem color, and eye color. Each solution includes the genotypes of the parents, the construction of the Punnett Square, and the resulting genotypes and phenotypes of the offspring, along with the probabilities of each outcome. The analysis extends to predicting the impact of gene dominance and crossing over on the expected ratios. This resource is designed to assist students in understanding and applying the principles of Mendelian genetics to solve inheritance problems.

Running head: PUNNETT SQUARE PROBLEM
PUNNETT SQUARE PROBLEM
Name of the Student
Name of the School
Author note

1
PUNNETT SQUARE PROBLEM
1.
a) Male Genotype: HH
Allele H and H
Mother genotype hh
Allele h, h
H H
h Hh Hh
h Hh Hh
b) There are 4 genotypes formed however all the genotypes are same Hh only
c) All the phenotypes will be long hair as they all have allele H that is dominant. Therefore, as
both the parents have homozygous allele the offerings will be heterozygous.

2
PUNNETT SQUARE PROBLEM
2)
R= dominant red flower allele, r= recessive white flower allele
R(dominant red) R
r Rr Rr
r (recessive white) Rr Rr
a) The genotype of parent will be RR and rr (as it is mentioned that both the parents are
homozygous dominant or recessive).
b) Parent genotype RR x rr (homozygous cross)
R R
r Rr (Red) Rr (Red)
r Rr (Red) Rr (Red)
All the offspring’s will be red as homozygous cross gives heterozygous genotype. Therefore, all
the offspring’s will be Rr, that means all flowers will be Red in colour. There will not be any
white flower as white is recessive allele here.

3
PUNNETT SQUARE PROBLEM
3.
a) Genotype for both parent will be Gg
b) Gg x Gg (heterozygous cross)
G g
G GG (25%) Gg (25%)
g Gg (25%) Gg (25%)
c) The probability of getting genotype Gg is 75%, as each Punnet block represent 25 % and Gg is
present in 3 blocks therefore the probability will be 75%.
d) As green is dominating the probability of getting green stem will be 75% as there as 75%
changes of getting plant with Gg genotype.
4.
a) The father genotype will be bb (as he is blue eyed)
b) Mother’s genotype will be either BB/ Bb (as she is brown eyed)

4
PUNNETT SQUARE PROBLEM
c) As the child is blue eyed, it means the child is having bb genotype. Therefore, the mother’s
genotype will be Bb
b b
B Bb (brown) Bb (brown)
b Bb (blue eyed) Bb (blue eyed)
5.
T t
T TT (25%) Tt (25%)
t Tt (25%) tt (25%)
a) Total plants genotype with tall phenotype will be TT or Tt and hence total number of tall
plants will be 1500
b) The genotype of offspring depends on crossing over and linkage. The gene dominance can be
predicted only during crossing over. Hence, if allele‘t’ becomes dominant during crossing over
then the number of tall plant will change.

5
PUNNETT SQUARE PROBLEM
6.
a) Parents phenotype will be
Father (Red eyed male) =Rr
Mother (White eyed female) = rr
b)
R r
r Rr (Red eyed) 25% rr (white eyed) 25%
r Rr (Red eyed) 25% rr (white eyed) 25%
c) Probability of offspring with white-eye will be 50%. The probability of getting genotype rr is
50%, as each Punnet block represent 25 % and rr is present in 2 blocks therefore the probability
will be 50%.