Tutorial 2 Chromosomes: Mapping & Sex determination
March 2024
Tutoriaal 2 Chromosome: Kartering & Geslagsbepaling
MEMO
1. In maize, sh produces shrunken endosperms in the kernels and Sh gives round kernels. The
recessive allele c gives endosperms with no pigment and C gives pigmented endosperms.
Two homozygotic plants are crossed. The F1 are all round and pigmented. The F1 is test
crossed and produces:
In mielies vorm sh verrimpelde endosperm in die pitte en Sh gee ronde pitte. Die resessiewe
alleel c lei tot kleurlose endosperm en C gee gekleurde endosperm. Twee homosigotiese
plante word gekruis. Die F1 was almal rond en gekleur. Die F1 word getoetskruis en vorm:
Shrunken, colourless | Verrimpeld, kleurloos 4035
Round, colourless | Rond, kleurloos 152
Round, pigmented | Rond, gekleur 4032
Shrunken, pigmented | Verrimpeld, gekleur 149
a. Determine with Chi-square analysis if the genes are linked. (HINT: Independent
assortment would give 1:1:1:1 ratio or in other words 50% parental types and 50%
recombinant types)
Bepaal met Chi-kwadraat analise of die gene gekoppeld is. (WENK: Onafhanklike
sortering sal 1:1:1:1 ratio lewer of anders gestel 50% ouerlike tipes en 50% rekombinante
tipes)
H0 = The genes are not linked and assort independently from each other.
Deviations from the expected 1:1:1:1 ratio is due to chance.
H0 = Die gene is nie gekoppeld nie en sorteer onafhanklik van mekaar. Afwykings
van die verwagte 1:1:1:1 verhouding is toevallig.
Observed Expected X2
4035 2092 1805
152 2092 1799
4032 2092 1799
149 2092 1805
Genetics | Genetika 214
1
, Tutorial 2 Chromosomes: Mapping & Sex determination
March 2024
Tutoriaal 2 Chromosome: Kartering & Geslagsbepaling
Total 8368 8368 7208
X 2 = 7028
DF = 3
P <<< 0.05
Conclusion: The genes do NOT assort independently. Differences between the
observed and expected ratios are NOT due to chance. Reject the null hypothesis.
Genes are therefore linked
Afleiding: Die gene skei nie onafhanklik nie. Verskille tussen waargenome en
verwagte getalle is NIE toevallig nie. Verwerp nil hipotese. Gene is dus gekoppeld
b. If the genes are linked, set up a chromosome linkage map for the two genes.
Indien die gene gekoppeld is, stel chromosoom koppeling kaart op vir die twee
gene.
Map distance = Number of recombinant offspring / total number offspring X 100
Kaartafstand = Aantal rekombinante nageslag / totale aantal nageslag X 100
(149 + 152) / (8368) X 100 = 3.60 cM
Sh 3.60 cM C
2. Assume that investigators crossed a strain of flies carrying the dominant eye mutation Lobe
on the second chromosome with a strain homozygous for the second chromosome
recessive mutations smooth abdomen and straw body. The F1 Lobe females were then
backcrossed with homozygous smooth abdomen, straw body males and the following
phenotypes were observed:
Neem aan data navorsers ʼn lyn van vlieë met die dominante oog mutasie “Lobe” op die
tweede chromosoom kruis met ʼn lyn homosigoties vir die tweede chromosoom resessiewe
mutasies “smooth abdomen” en “straw body”. Die F1 Lob wyfies word dan teruggekruis
met homosigotiese gladde maag, strooikleurige mannetjies en die volgende fenotipes
word waargeneem:
smooth abdomen, straw body 820
smooth abdomen, Lobe 42
Lobe, straw body 152
straw body 58
smooth abdomen 148
Genetics | Genetika 214
2
March 2024
Tutoriaal 2 Chromosome: Kartering & Geslagsbepaling
MEMO
1. In maize, sh produces shrunken endosperms in the kernels and Sh gives round kernels. The
recessive allele c gives endosperms with no pigment and C gives pigmented endosperms.
Two homozygotic plants are crossed. The F1 are all round and pigmented. The F1 is test
crossed and produces:
In mielies vorm sh verrimpelde endosperm in die pitte en Sh gee ronde pitte. Die resessiewe
alleel c lei tot kleurlose endosperm en C gee gekleurde endosperm. Twee homosigotiese
plante word gekruis. Die F1 was almal rond en gekleur. Die F1 word getoetskruis en vorm:
Shrunken, colourless | Verrimpeld, kleurloos 4035
Round, colourless | Rond, kleurloos 152
Round, pigmented | Rond, gekleur 4032
Shrunken, pigmented | Verrimpeld, gekleur 149
a. Determine with Chi-square analysis if the genes are linked. (HINT: Independent
assortment would give 1:1:1:1 ratio or in other words 50% parental types and 50%
recombinant types)
Bepaal met Chi-kwadraat analise of die gene gekoppeld is. (WENK: Onafhanklike
sortering sal 1:1:1:1 ratio lewer of anders gestel 50% ouerlike tipes en 50% rekombinante
tipes)
H0 = The genes are not linked and assort independently from each other.
Deviations from the expected 1:1:1:1 ratio is due to chance.
H0 = Die gene is nie gekoppeld nie en sorteer onafhanklik van mekaar. Afwykings
van die verwagte 1:1:1:1 verhouding is toevallig.
Observed Expected X2
4035 2092 1805
152 2092 1799
4032 2092 1799
149 2092 1805
Genetics | Genetika 214
1
, Tutorial 2 Chromosomes: Mapping & Sex determination
March 2024
Tutoriaal 2 Chromosome: Kartering & Geslagsbepaling
Total 8368 8368 7208
X 2 = 7028
DF = 3
P <<< 0.05
Conclusion: The genes do NOT assort independently. Differences between the
observed and expected ratios are NOT due to chance. Reject the null hypothesis.
Genes are therefore linked
Afleiding: Die gene skei nie onafhanklik nie. Verskille tussen waargenome en
verwagte getalle is NIE toevallig nie. Verwerp nil hipotese. Gene is dus gekoppeld
b. If the genes are linked, set up a chromosome linkage map for the two genes.
Indien die gene gekoppeld is, stel chromosoom koppeling kaart op vir die twee
gene.
Map distance = Number of recombinant offspring / total number offspring X 100
Kaartafstand = Aantal rekombinante nageslag / totale aantal nageslag X 100
(149 + 152) / (8368) X 100 = 3.60 cM
Sh 3.60 cM C
2. Assume that investigators crossed a strain of flies carrying the dominant eye mutation Lobe
on the second chromosome with a strain homozygous for the second chromosome
recessive mutations smooth abdomen and straw body. The F1 Lobe females were then
backcrossed with homozygous smooth abdomen, straw body males and the following
phenotypes were observed:
Neem aan data navorsers ʼn lyn van vlieë met die dominante oog mutasie “Lobe” op die
tweede chromosoom kruis met ʼn lyn homosigoties vir die tweede chromosoom resessiewe
mutasies “smooth abdomen” en “straw body”. Die F1 Lob wyfies word dan teruggekruis
met homosigotiese gladde maag, strooikleurige mannetjies en die volgende fenotipes
word waargeneem:
smooth abdomen, straw body 820
smooth abdomen, Lobe 42
Lobe, straw body 152
straw body 58
smooth abdomen 148
Genetics | Genetika 214
2
