Tutorial 3 Chromosome Mutations & Integrated Concepts
17 Apr. 2024
Tutoriaal 3 Chromosomeale Mutasies & Geïntegreerde Konsepte
MEMO
1. According to Mendel’s law of segregation, alleles A and a segregate from each other and
appear in equal numbers among the gametes. But Mendel did not know that his plants
were diploid. In fact, because plants are frequently tetraploid, he could have been
unlucky enough to have started with peas that were 4n rather than 2n. Let us assume that
Mendel’s peas were tetraploid, that every gamete contains two alleles, and that the
distribution of alleles to the gamete is random. Suppose we have a cross of AAAA x aaaa
where A is dominant, regardless of the number of a alleles present in an individual.
Volgens Mendel se wet van segregasie sal allele A en a van mekaar skei en in gelyke
hoeveelhede verskyn in die gamete. Maar Mendel het nie geweet dat sy plante diploïed
was nie. In werklikheid aangesien plante gereeld tetraploïed is, kon hy ongelukkig genoeg
gewees het om met 4n ertjieplante te werk eerder as 2n plante. Kom ons neem aan dat
Mendel se ertjieplante tetraploïed was, dat elke gameet twee allele bevat het en dat die
verspreiding van allele in die gamete willekeurig was. Veronderstel jy kruis AAAA x aaaa
waar A dominant is ongeag van die hoeveelheid a allele wat teenwoordig is in ’n individu.
a. What will be the genotype/s of the F1 peas?
Wat sal die genotipe/s wees van die F1 ertjies?
AAaa
b. If the F1 peas are selfed, what will be the phenotypic ratios in the F2 peas?
Indien die F1 ertjies selfbestuif word, wat sal die fenotipiese ratio’s wees in die F2
ertjies?
A1A2a1a2 x A1A2a1a2
Six possible gametes
A1A2; A1a1; A1a2; A2a1; A2a2; a1a2
Genetics | Genetika 214
1
,Tutorial 3 Chromosome Mutations & Integrated Concepts
17 Apr. 2024
Tutoriaal 3 Chromosomeale Mutasies & Geïntegreerde Konsepte
Thus,
1/6 AA 4/6 Aa 1/6 aa
1/6 AA 1/36 AAAA 4/36 AAAa 1/36 AAaa
4/6 Aa 4/36 AAAa 16/36 AAaa 4/36 Aaaa
1/6 aa 1/36 AAaa 4/36 Aaaa 1/36 aaaa
Phenotypic ratio: 35 A-: 1 aa
2. Eyeless is a recessive allele (ey) on the fourth chromosome of Drosophila. A male trisomic
for chromosome IV with the genotype Ey/ey/ey is crossed to a normal diploid eyeless
female. What expected genotypic and phenotypic ratios would be found in the resultant
offspring from the random assortment of the chromosomes to the gametes?
Oogloos is ʼn resessiewe alleel (ey) op die vierde chromosoom van Drosophila. ʼn Mannetjie
wat trisomies is vir chromosoom IV met die genotipe +/ey/ey word gekruis met ʼn normale
diploïede ooglose wyfie. Watter verwagte genotipiese en fenotipiese ratio’s sal ontstaan
in die gevolglike nageslag as gevolg van die willekeurige sortering van chromosome na
die gamete?
Ey ey1 ey2 X ey ey
(The superscripts are used just as a means of keeping track with the similar alleles in the
male)
Six possible male gametes
Ey ey1; ey2; EY ey2; ey1; Ey; ey1 ey2
GENOTYPIC
Ey ey ey (2/6) [normal eyes but trisomic]
ey ey (2/6) [eyeless]
Ey ey (1/6) [normal eyes]
ey ey ey (1/6) [eyeless but trisomic]
PHENOTYPIC
3/6 normal (one disomic and two trisomic)
3/6 eyeless (two disomic and one trisomic)
Genetics | Genetika 214
2
, Tutorial 3 Chromosome Mutations & Integrated Concepts
17 Apr. 2024
Tutoriaal 3 Chromosomeale Mutasies & Geïntegreerde Konsepte
3. The New World cotton species Gossypium hirsutum has a 2n chromosome number of 52.
The Old World species G. thurberi and G. herbaceum each have a 2n number of 26.
Hybrids between these species show the following chromosome pairing arrangements
during meiosis:
Die Nuwe Wêreld katoenspesie Gossypium hirsutum het ʼn 2n chromosoom getal van 52.
Die Ou Wêreld spesies G. thurberi en G. herbaceum het elk ʼn 2n getal van 26. Basters
tussen hierdie spesies toon die volgende chromosoomparingsrangskikkings tydens meiose:
Hybrid Pairing arrangement
Baster Paringsrangskikking
G. hirsutum × G. thurberi 13 small bivalents + 13 large univalents
13 klein bivalente + 13 groot univalente
G. hirsutum × G. herbaceum 13 large bivalents + 13 small univalents
13 groot bivalente + 13 klein univalente
G. thurberi × G. herbaceum 13 large univalents + 13 small univalents
13 groot univalente + 13 klein univalente
Interpret these observations to clearly indicate the relationships between the New World and
two Old World species and also indicate the chromosome composition (small or large or
both) for these three species.
Interpreteer hierdie waarnemings om duidelik die verwantskap tussen die Nuwe Wêreld en
twee Ou Wêreld spesies aan te dui en dui ook die chromosoom samestelling (klein of groot
of beide) aan vir hierdie drie spesies.
hirsutum and thurberi share some similarity in their chromosomes (forming Small
bivalents)
Similarly, hirsutum and hebaceum share some similarity in their chromosomes (forming
Large bivalents)
thurberi and herbaceum do not share similar chromosomes (only univalents are
present)
It can de deduced that herbaceum and thurberi hybridised and subsequently
poliploidization occurred to give rise to hirsutum
G. hirsutum 26 large + 26 small
G. thurberi 26 small
G. herbaceum 26 large
Genetics | Genetika 214
3
17 Apr. 2024
Tutoriaal 3 Chromosomeale Mutasies & Geïntegreerde Konsepte
MEMO
1. According to Mendel’s law of segregation, alleles A and a segregate from each other and
appear in equal numbers among the gametes. But Mendel did not know that his plants
were diploid. In fact, because plants are frequently tetraploid, he could have been
unlucky enough to have started with peas that were 4n rather than 2n. Let us assume that
Mendel’s peas were tetraploid, that every gamete contains two alleles, and that the
distribution of alleles to the gamete is random. Suppose we have a cross of AAAA x aaaa
where A is dominant, regardless of the number of a alleles present in an individual.
Volgens Mendel se wet van segregasie sal allele A en a van mekaar skei en in gelyke
hoeveelhede verskyn in die gamete. Maar Mendel het nie geweet dat sy plante diploïed
was nie. In werklikheid aangesien plante gereeld tetraploïed is, kon hy ongelukkig genoeg
gewees het om met 4n ertjieplante te werk eerder as 2n plante. Kom ons neem aan dat
Mendel se ertjieplante tetraploïed was, dat elke gameet twee allele bevat het en dat die
verspreiding van allele in die gamete willekeurig was. Veronderstel jy kruis AAAA x aaaa
waar A dominant is ongeag van die hoeveelheid a allele wat teenwoordig is in ’n individu.
a. What will be the genotype/s of the F1 peas?
Wat sal die genotipe/s wees van die F1 ertjies?
AAaa
b. If the F1 peas are selfed, what will be the phenotypic ratios in the F2 peas?
Indien die F1 ertjies selfbestuif word, wat sal die fenotipiese ratio’s wees in die F2
ertjies?
A1A2a1a2 x A1A2a1a2
Six possible gametes
A1A2; A1a1; A1a2; A2a1; A2a2; a1a2
Genetics | Genetika 214
1
,Tutorial 3 Chromosome Mutations & Integrated Concepts
17 Apr. 2024
Tutoriaal 3 Chromosomeale Mutasies & Geïntegreerde Konsepte
Thus,
1/6 AA 4/6 Aa 1/6 aa
1/6 AA 1/36 AAAA 4/36 AAAa 1/36 AAaa
4/6 Aa 4/36 AAAa 16/36 AAaa 4/36 Aaaa
1/6 aa 1/36 AAaa 4/36 Aaaa 1/36 aaaa
Phenotypic ratio: 35 A-: 1 aa
2. Eyeless is a recessive allele (ey) on the fourth chromosome of Drosophila. A male trisomic
for chromosome IV with the genotype Ey/ey/ey is crossed to a normal diploid eyeless
female. What expected genotypic and phenotypic ratios would be found in the resultant
offspring from the random assortment of the chromosomes to the gametes?
Oogloos is ʼn resessiewe alleel (ey) op die vierde chromosoom van Drosophila. ʼn Mannetjie
wat trisomies is vir chromosoom IV met die genotipe +/ey/ey word gekruis met ʼn normale
diploïede ooglose wyfie. Watter verwagte genotipiese en fenotipiese ratio’s sal ontstaan
in die gevolglike nageslag as gevolg van die willekeurige sortering van chromosome na
die gamete?
Ey ey1 ey2 X ey ey
(The superscripts are used just as a means of keeping track with the similar alleles in the
male)
Six possible male gametes
Ey ey1; ey2; EY ey2; ey1; Ey; ey1 ey2
GENOTYPIC
Ey ey ey (2/6) [normal eyes but trisomic]
ey ey (2/6) [eyeless]
Ey ey (1/6) [normal eyes]
ey ey ey (1/6) [eyeless but trisomic]
PHENOTYPIC
3/6 normal (one disomic and two trisomic)
3/6 eyeless (two disomic and one trisomic)
Genetics | Genetika 214
2
, Tutorial 3 Chromosome Mutations & Integrated Concepts
17 Apr. 2024
Tutoriaal 3 Chromosomeale Mutasies & Geïntegreerde Konsepte
3. The New World cotton species Gossypium hirsutum has a 2n chromosome number of 52.
The Old World species G. thurberi and G. herbaceum each have a 2n number of 26.
Hybrids between these species show the following chromosome pairing arrangements
during meiosis:
Die Nuwe Wêreld katoenspesie Gossypium hirsutum het ʼn 2n chromosoom getal van 52.
Die Ou Wêreld spesies G. thurberi en G. herbaceum het elk ʼn 2n getal van 26. Basters
tussen hierdie spesies toon die volgende chromosoomparingsrangskikkings tydens meiose:
Hybrid Pairing arrangement
Baster Paringsrangskikking
G. hirsutum × G. thurberi 13 small bivalents + 13 large univalents
13 klein bivalente + 13 groot univalente
G. hirsutum × G. herbaceum 13 large bivalents + 13 small univalents
13 groot bivalente + 13 klein univalente
G. thurberi × G. herbaceum 13 large univalents + 13 small univalents
13 groot univalente + 13 klein univalente
Interpret these observations to clearly indicate the relationships between the New World and
two Old World species and also indicate the chromosome composition (small or large or
both) for these three species.
Interpreteer hierdie waarnemings om duidelik die verwantskap tussen die Nuwe Wêreld en
twee Ou Wêreld spesies aan te dui en dui ook die chromosoom samestelling (klein of groot
of beide) aan vir hierdie drie spesies.
hirsutum and thurberi share some similarity in their chromosomes (forming Small
bivalents)
Similarly, hirsutum and hebaceum share some similarity in their chromosomes (forming
Large bivalents)
thurberi and herbaceum do not share similar chromosomes (only univalents are
present)
It can de deduced that herbaceum and thurberi hybridised and subsequently
poliploidization occurred to give rise to hirsutum
G. hirsutum 26 large + 26 small
G. thurberi 26 small
G. herbaceum 26 large
Genetics | Genetika 214
3
