Genetics and Inheritance: Biology Assignment - Problem Solutions

This document presents solutions to a genetics assignment, addressing multiple choice questions and problem-solving scenarios. The assignment covers fundamental concepts in genetics, including the study of how genes interact to produce a phenotype, inheritance patterns, and the application of Punnett squares. Solutions are provided for problems involving chicken comb traits, blood types, hair and curly hair traits, flower color inheritance, and color blindness. The answers are explained step-by-step, including calculations of genotypic and phenotypic ratios. Furthermore, the assignment explores the mechanisms of genetic diversity arising from meiosis and sexual reproduction, such as crossing over and independent assortment, and the advantages of genetic diversity within a population. References are included to support the answers provided. This document serves as a comprehensive resource for understanding and solving genetics problems.

Running head: GENETIC DIVERSITY
GENETIC DIVERSITY
Name of Student:
Name of University:
Author’s Note:

1GENETIC DIVERSITY
Answer no 1.
A) The study of how gene interact to produce a phenotype
Answer no 2.
True
Answer no 3.
As rose combed chicken is dominant to single combed chicken,
Allele of Rose combed chicken – RR (homozygous dominant)
Allele of single-combed chicken- rr (homozygous recessive)
When homozygous rose combed chicken (RR) mated with the single combed hen (rr), all the
chicken in the F1 generation will be heterozygous rose combed.
Homozygous rose combed chicken
R R
Recessive single
combed Hen
r Rr Rr
r Rr Rr
Therefore, in F1 generation, all the chicken have Rr gametes and are heterozygous rose combed
chicken.

2GENETIC DIVERSITY
In F2, there is self-mating of Rr × Rr
Heterozygous rose combed chicken
R r
Heterozygous rose
combed Hen
R RR
(rose combed)
Rr
rose combed
r Rr
(rose combed)
rr
(single-combed)
A) The genotypic ratio of F2 generation is 1(RR) :2 (Rr):1 (rr)
B) The phenotypic ratio of F2 generation is 3 (red combed) : 1 (single-combed)
Answer no 4.
Allele of heterozygous man for blood type A – IA Io
Allele of Woman heterozygous for blood type B- IB Io
IA Io × IB Io
In order to have type O blood group, both of the parents must carry a recessive allele.

3GENETIC DIVERSITY
Blood type A male
IA Io
Blood type B female IB IA IB
blood type AB 1/4
IB IO
Blood type B 1/4
Io IA IO
Blood type A 1/4
IO IO
Blood type O 1/4
Therefore, out of 4 genotype, 1 genotype code for O blood type. The probability of occurrence of
blood type O child is ¼.
Answer no 5.
Dark hair is dominant (D) and blonde hair is recessive (d)
Curly hair is dominant (C) and straight hair is recessive (c).
Genotype of women having blonde and curly hair but heterozygous – ddCc
Genotype of man with straight and dark hair but heterozygous- Ddcc
Gametes for women ddCc

4GENETIC DIVERSITY
The gametes are Cd, cd, Cd, cd
Gametes for men Ddcc
The gametes are Dc, dc, Dc, dc.
The dihybrid cross:
dC dc dC dc
Dc DdCc
Dark and curly
hair
Ddcc
Dark and
straight hair
DdCc
Dark and curly
hair
Ddcc
Dark and
straight hair
dc ddCc
blonde and
curly hair
ddcc
blonde and
straight hair
ddCc
blonde and
curly hair
ddcc
blonde and
straight hair
Dc DdCc
Dark and curly
Ddcc
Dark and
DdCc
Dark and curly
Ddcc
Dark and
d d cC
D d cc

5GENETIC DIVERSITY
hair straight hair hair straight hair
dc ddCc
blonde and
curly hair
ddcc
blonde and
straight hair
ddCc
blonde and
curly hair
ddcc
blonde and
straight hair
The phenotypic ratio is 4 (DdCc): 4 (Ddcc): 4 (ddCc): 4(ddcc)
4 (Dark and curly hair): 4(Dark and straight hair): 4 (blonde and curly hair): 4(blonde and
straight hair).
Answer no 6.
a) Allele of blue flower – BB
Allele for yellow flower YY
B for blue and Y for yellow
Crossing of blue and yellow BB × YY
B B
Y BY (blue with
yellow)
BY (blue with
yellow)
Y BY (blue with
yellow)
BY (blue with
yellow)
The genotypic ratio is 1 (BY): 1 (BY): 1 (BY) : 1(BY)

6GENETIC DIVERSITY
Phenotypic ratio is 1 (blue and yellow flower) all the flower were yellow with blue streaks.
b) Cross between two yellow flowers with blue streaks
BY× BY
B Y
B BB (blue) BY (blue with
yellow)
Y BY (blue with
yellow)
YY (yellow)
The genotypic ratio is 1 (BB): 2 (BY): 1 (YY)
Phenotypic ration is 1 (blue flower):2 (blue and yellow flower): 1 (yellow flower).
Answer no. 7
a) Cross between Red Green color blindness male and carrier female ,Genotype of Red
Green- color blindness male- XC Y, Genotype of carrier female- XC Xc
C –color blind, and c- normal
XC Y
XC XC XC XC Y
Xc XC Xc Xc Y
Proportion of color blind male (XCY) is 1 out of 4 which is estimated to 25%

7GENETIC DIVERSITY
Proportion of color blind female (XC XC) which is 1 out of 4, hence is estimated to be 25%
b) Cross between normal male and RGCB female
Xc Y × XC XC
Xc Y
XC XC Xc
Carrier
female
XC Y
Affected male
1/2
XC XC Xc
Carrier
female
XC Y
Affected male
1/2
As seen from the cross half portion of male are color blind (XC Y) and none of the female is
color blind.
Answer no 8.
The two ways by which meiosis and sexual reproduction create genetic diversity are
 Crossing over- In meiosis, homologous chromosome pair together and exchange their
part of segment with each other. This exchange of gene result in genetic diversity (Li et
al. 2018)
 Independent assortment- In this process, each pair of gene get independently assorted
and each chromosome shuffles and gets into the separate gamete resulting in genetic
diversity (Bartee, 2017).

8GENETIC DIVERSITY
Genetic diversity is advantageous because it allow the population to adapt with the changing
environment and contributes to evolution of new gene and allele ((Li et al. 2018).
Answer no 9.
2 males and 1 females are affected.

9GENETIC DIVERSITY
Reference
Bartee, L., 2017. Meiosis I. Principles of Biology: Biology 211, 212, and 213.
Li, R., Bitoun, E., Altemose, N., Davies, R.W., Davies, B. and Myers, S.R., 2018. A high-
resolution map of non-crossover events in mice reveals impacts of genetic diversity on meiotic
recombination. bioRxiv, p.428987.