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]