7 Answers to end-of-chapter questions
1 D [1]
2 C [1]
3 B [1]
4 B [1]
5 C [1]
6 B [1]
7 A [1]
8 D [1]
9 C [1]
10 D [1]
Structured questions 11 a Dihybrid
2 traits were studied / colour of seeds and seed shape [1] b 9:3:3:1 [1] c
Smooth Wrinkled 34 points [3]Number of seeds 315 + 108 = 423 101 + 32 = 133 2 points [2]Phenotypic Ratio 3.2 1 1 point [1] Yellow Green 34 points [3]Number of seeds 315 + 101 = 416 108 + 32 = 140 2 points [2]Phenotypic Ratio 2.97 1 1 point [1]
d The monohybrid ratio was obtained for each trait / 3:1 ratio
The two traits were behaving as though they were separate from each other Any 2 points [2]
e No. [1]
Epistasis: one gene may influence the expression of another gene Both alleles may be on same chromosome One trait or both may be sex-linked / carried on X chromosome [1]
12 a The presence of a dominant allele at one locus prevents the expression of alleles
at another Well explained [2]
b WwYy white Wwyy white wwYY yellow 4 correct [2] wwyy green 23 correct [1] c i
Parental phenotypes white green Parental genotypes WWYY wwyy Gametes
F1 genotypes all WwYy F1 phenotypes all white
F1 are interbred to give an F2 generation
Parental genotypes
WwYy WwYy
Gametes
or or or
or or or
Gametes from one parent
WWYY white
WWYy white
WwYY white
WwYy white
WWYy white
WWyy white
WwYy white
Wwyy white
WwYY white
WwYy white
wwYY yellow
wwYy yellow
Gam
etes
from
oth
er p
aren
t
WwYy white
Wwyy white
wwYy yellow
wwyy green
Both F1 and F2 correct [4]F1 correct [1]F2 correct [3]Partly correct
from F1 and F2 [2]
[max 4] ii 12 white : 3 yellow : 1 green [1]
WY wy
WY Wy
wY wy
WY Wy
wY wy
WY
WY
Wy
Wy wY
wY
wy
wy
d
Scallop fruits Pear fruits Observed results [O] 76 34
Expected ratio 3 1 Expected results [E] 82.5 27.5
O E 6.5 6.5 [O E]2 42.25 42.25 [O E]2
E 0.51 1.54
2 = [O E]2 = 2.05 E
All correct [4]810 points [3]
57 points [2]24 points [1]
[max 4]
e There is no significant difference between the observed and expected results [1]
2 calculated value (2.05) for 1 degree of freedom is less than 3.84 (2 table) at the critical value of p = 0.05. [1]
13 a i Traits which can be shown by both sexes but whose inheritance is
linked with the inheritance of the sex chromosomes [1] The X chromosome is longer than the Y chromosome, it has many
genes not found on the Y [1]
ii A form of the gene which only has an effect on the phenotype when the dominant allele is absent / influences the phenotype only in the presence of another identical allele [1]
b It is linked to the inheritance of the sex / X chromosome [1]
Mostly present in the male offspring [1] If mother has the trait for colour blindness ; it is passed to the male
offspring [1] c Dominant/recessive inheritance [1]
Most individuals not colour blind Individuals with both alleles / heterozygous are not colour blind (e.g. individual 6) Colour blindness is only reflected in the phenotype if both alleles are
identical / dominant allele is not present Any other point [1] d
2 XnXn 3 XnY 7 XnY 11 XNXn 13 XnY 19 XNY
e 0.5 / / 50%
Mother is heterozygous / XNXn / carrier for the trait
Since the mother passes the X chromosome to her son, she can therefore pass either XN or Xn Can be shown in a genetic diagram
Parental phenotypes
male, colour blind female, normal vision
Parental genotypes
XnY XNXn
Gametes
F1 genotypes Xn Y XN Y XN Xn XnXn
F1 phenotypes
1 colour-blind : 1 normal : 1 normal : 1 colour-blind male male female female
Correct probability [1]
Explanation [3]
Xn XN Y Xn
Essay questions 14 a i Monohybrid inheritance / monohybrid ratio
Cross between 2 heterozygous individuals E.g. if round pea is dominant (R) over wrinkled pea (r), then the heterozygotes genotype
is Rr
Parental phenotypes
round round
Parental genotypes
Rr Rr
Gametes
F1 genotypes RR Rr Rr rr F1 phenotypes 3 round : 1 wrinkled Results in 3 round : 1 wrinkled ratio
Well explained [2]Information can be
obtained from genetic diagram
ii Monohybrid test cross ratio
Cross between a heterozygous individual and a recessive individual For example: if one wanted to determine if the round seeds above were heterozygous or
homozygous for the trait, a test cross would be done
Parental phenotypes
round wrinkled
Parental genotypes
Rr rr
Gametes
F1 genotypes Rr rr F1 phenotypes 1 round : 1 wrinkled Results in 1 round : 1 wrinkled ratio
Well explained [2]Information can be
obtained from genetic diagram
iii Codominance: cross between two heterozygotes
Both alleles make a contribution to the phenotype, e.g. red (R) and white (W) hair colour in cows
Heterozygote is phenotypically different from homozygotes Ratio occurs when two heterozygotes are crossed, e.g. a roan coloured cow
and bull
R r R r
R r r
Parental phenotypes
roan roan
Parental genotypes
RW RW
Gametes
F1 genotypes RR RW RW WW F1 phenotypes 1 red : 2 roan : 1 white
Well explained [2]Information can be
obtained from genetic diagram
Results in 1 red: 2 roan: 1 white iv Dihybrid cross between 2 heterozygous individuals / dihybrid ratio
Dominant, recessive Inheritance of 2 traits at a time Black hair colour is dominant to brown hair colour in guinea pigs; and short
hair is dominant to long hair e.g.
Guinea pig female Gametes BS Bs bS bs
BS BBSS black short
BBSs black short
BbSS black short
BbSs black short
Bs BBSs black short
BBss black long
BbSs black short
Bbss black long
bS BbSS black short
BbSs black short
bbSS brown short
bbSs brown short
Gui
nea
pig
mal
e
bs BbSs black short
Bbss black long
bbSs brown short
bbss brown long
Results in 9 black short : 3 black long : 3 brown short : 1 brown long
Well explained [2]Information can be
obtained from genetic diagram
R W R W
v Dihybrid test cross ratio Cross between a heterozygous individual and a double recessive individual Used to determine the genotype of the heterozygote Smooth pea is dominant over wrinkled, yellow is dominant over green
Gametes from one parent
SsYy
Ssyy
ssYy
ssyy
Gam
etes
from
oth
er
pare
nt
1 smooth yellow
1 smooth green
1 wrinkled yellow
1 wrinkled
green
Well explained [2]Information can be
obtained from genetic diagram
b Determined by an autosomal gene
Gene has 3 forms / A, B, O Allele A codes for Antigen A Allele B codes for Antigen B Allele O does not code for any antigen A and B are dominant to O / O is recessive to A and B A and B are codominant Possible genotypes and blood groups are:
Genotypes Blood groups
IAIA
IAIO
IBIB
IBIO
IAIB
IOIO
A A B B
AB O
56 pts [3]34 pts [2] 12 pts [1]
c A [1]
B [1] 15 a i Alleles of a gene that are both fully expressed in the phenotype of the offspring.
Neither allele is dominant nor recessive to each. For example, in the human ABO blood group system, alleles A and B are codominant with each other, so a person with both has blood type AB Well explained [1]
ii An organism that has the same alleles at a particular gene locus on homologous
chromosomes Well explained [1] iii A genotype in which the alleles of a gene are different / diploid condition in
which alleles at a given locus are different Well explained [1] iv Pertaining to chromosomes not considered to be the sex chromosomes Well explained [1]
sy
SY Sy sY sy
b
Parental phenotypes
black frizzled grey mildly frizzled
Parental genotypes
CBCB SFSF
CBCW SSSF
Gametes
F1 genotypes CBCB SSSF CBCB SFSF CBCW SSSF CBCW SFSF
F1 phenotypes
1 black : 1 black : 1 grey : 1 grey mildly frizzled mildly frizzled frizzled frizzled
Each row [1][max 5]
c No [1]
Phenotypes are different for homozygotes and heterozygotes [1] d Cross a black frizzled rooster with a white straight feathered hen / reverse
Heterozygotes would have intermediate traits Crossing purebred chickens for contrasting traits would produce the heterozygotes
Parental phenotypes black frizzled white straight Parental genotypes
CBCB SFSF
CWCW SSSS Gametes
F1 genotypes CBCW SSSF
F1 phenotypes all grey, mildly frizzled
Well explained with genetic diagram [4]
Any 1 point [1][max 4]
16 a To determine if the differences between the observed and expected results are
significant or insignificant To estimate the probability of the differences of observed and expected results
being due to chance To see if the observed results fit the expected genetic ratio Any point [2]
b i Allele: alternative forms of same gene responsible for contrasting traits [1]
Gene: length of DNA / sequence of nucleotides which codes for polypeptide [1] ii The observed results are not significantly different from the expected results [2] iii Value in 2 table at p = 0.05 under the appropriate degrees of freedom (number
of classes 1) Well explained [2] iv 1:1:1:1 [1]
CBSF CBSS CBSF CWSS CWSF
CBSF CWSS
v
Rough black
Rough white
Smooth black
Smooth white
Observed results [O] 22 18 25 19 Expected ratio 1 1 1 1
Expected results [E] 21 21 21 21 O E 1 -3 4 -2
[O E]2 1 9 16 4 [O E]2
E 0.05 0.43 0.76 0.19
2 = [O E]2 = 1.43 E
Degrees of freedom = number of classes 1 = 4 1 = 3 Critical value is at p = 0.05, df = 3 in chi-square table Conclusion: if the 2 calculated value is less than the 2 value at p = 0.05,
df = 3, then the null hypothesis is true and the differences are insignificant / or reverse
Correct table [5]Each point to reach
conclusion [1][max 8]
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