Tutorial 4 Quantitative- & Population Genetics
May 2024
Tutoriaal 4 Kwantitatiewe- & Populasie Genetika
MEMO
1. A goldfish breeder is interested in two traits, tail-fin size (larger elaborate fins are deemed
more desirable) and resistance to fin rot (a fungal disease). His current breeding
population has an average fin length of 2 cm and on average fish take 7 days to recover
from infection. He runs a breeding experiment and selects fish with a mean tail fin length
of 4 cm and that have a mean recovery time of 3 days; the resulting offspring has a
mean tail fin length of 3.6 cm and a recovery time of 6.8 days. Which trait will be
improved by a continued selective breeding programme and why? Motivate with
appropriate calculations.
ʼn Goudvisteler stel belang in twee eienskappe, stertvingrootte (groter uitgebreide vinne
is meer gesog) en weerstand teen vinvrot ('n swamsiekte). Sy huidige broeipopulasie het
'n gemiddelde vinlengte van 2 cm en gemiddeld neem ʼn vis 7 dae om van infeksie te
herstel. Hy voer 'n teeleksperiment uit en kies vis met 'n gemiddelde vinlengte van 4 cm
en 'n gemiddelde hersteltyd van 3 dae; die gevolglike nageslag het ʼn gemiddelde
vinlengte van 3.6 cm en 6.8 dae herstel tyd. Watter eienskap sal verbeter word deur 'n
voortgesette selektiewe teelprogram en hoekom? Motiveer met toepaslike
berekeninge.
Tail-fin length | Stertvinlengte: Fin rot resistance | Vinvrotweerstand
h2 = (M2-M)/(M1-M) h2 = (M2-M)/(M1-M)
= (3.6-2)/ (4-2) = (6.8-7)/ (3-7)
= 0.8 = 0.05
Tail-fin-length seems to give a response to selection and this is supported by the realised
heritability estimate that is closer to one. On the contrary fin rot resistance has a realised
heritability closer to zero.
Stertvinlengte blyk ʼn respons op seleksie te gee en dit word ondersteun deur die
gerealiseerde oorerflikheidsberaming wat nader aan een is. Inteendeel, die
vinvrotweerstand het 'n gerealiseerde erflikheid van nader aan nul.
Genetics | Genetika 214
1
,Tutorial 4 Quantitative- & Population Genetics
May 2024
Tutoriaal 4 Kwantitatiewe- & Populasie Genetika
2. A researcher wants to investigate how much the environment influences stalk diameter
in ‘n wild strain of sugarcane. She grows the wild strain plants, with the cultivar N17 (highly
inbred) as an experimental control, in a field on the research farm. The phenotypic
variance for stalk diameter in the wild strain is 6.3 cm2, whilst the phenotypic variance for
N17 is only 2.1 cm2. What is the genotypic variance and the broad sense heritability in
the wild strain and what can be deduced about the influence of the environment on
phenotypic expression?
ʼn Navorser wil bepaal hoeveel die stingeldeursneë in ʼn wilde lyn suikerriet beïnvloed
word deur die omgewing. Sy groei plante van die wilde lyn met die van kultivar N17
(hoogs ingeteel), as ʼn eksperimentele kontrole, in die veld op die proefplaas. Die
fenotipiese variansie vir stingeldeursneë in die wilde lyn is 6.3 cm2, terwyl die fenotipiese
variansie vir N17 slegs 2.1 cm2 is. Wat is die genotipiese variansie en breë sin oorerflikheid
van die wilde lyn en wat kan afgelei word m.b.t. die omgewingsinvloed op fenotipiese
uitdrukking?
VP(wild) = 6.3 cm2
VP(inbred) = 2.1, in this line VG=0 (highly inbred), therefore VP = VE = 2.1
VP(wild) = VG + VE = VG + 2.1 = 6.3, therefore VG = 4.2 cm2
H2 = VG / VP = 4..3 = 0.67
Genetic differences between wild plants accounts for 67% of the observed variance in
phenotypic expression (broadsense heretibility). Thus the environment accounts for
33%.
Genetiese verskille tussen wilde plante verklaar 67% van die waargenome variansie in
fenotipiese uitdrukking (breë-sin oorerflikheid). Dus is die omgewing is verantwoordelik
vir 33%.
3. In a large flock of 5468 sheep, 76 animals have yellow fat, compared to the rest of the
members of the flock, which have white fat. It is assumed that this population is in HW
equilibrium for this autosomal locus. Furthermore, when sheep that are pure breeding for
white fat is mated to sheep that are pure breeding for yellow fat all of the resulting
offspring have white fat. When these F1’s are allowed to breed randomly, 25% of their
offspring have yellow fat.
In ʼn groot kudde van 5468 skape, het 76 diere geel vet, in vergelyking met die res van
die kudde wat wit vet het. Dit word aangeneem dat die populasie in HW ewewig is vir
hierdie outosomale lokus? Verder, wanneer suiwertelende wit vet skape geteel word
Genetics | Genetika 214
2
, Tutorial 4 Quantitative- & Population Genetics
May 2024
Tutoriaal 4 Kwantitatiewe- & Populasie Genetika
met suiwertelende geel vet skape, het al die nageslag wit vet. Wanneer hierdie F1’s
onderling lukraak teel, het 25% van hul nageslag geel vet.
a. What are the frequencies of the yellow- and white alleles in this population?
Wat is die frekwensie van die geel -en wit allele in die populasie?
Important: If both homozygous and heterozygous genotypic frequencies are given, the
formula {p = fr (p2) + ½ fr (2pq)} etc. must be used. You can only use the square root to
determine allele frequency if you cannot distinguish between the dominant
homozygotes and the heterozygotes.
Belangrik: Indien beide homosigotiese en heterosigotiese genotipiese frekwensies
gegee word moet jy {p = fr (p2) + ½ fr (2pq)} ens. gebruik om alleelfrekwensies te
bereken. Jy kan slegs die vierkantswortel gebruik om ‘n alleelfrekwensie te bepaal
indien daar nie onderskeid getref kan word tussen die dominante homosigote en die
heterosigote nie.
W = white fat | wit vet ; w = yellow fat | geel vet
Fr(WW) = p2
Fr(Ww) = 2pq
Fr(ww) = q2
q2 = = 0.014
q = √0.014
q = 0.12
p = 1-q = 0.88
b. What percentage of sheep with white fat are heterozygous carriers of the yellow fat
allele?
Watter persentasie van skape met wit vet is heterosigotiese draers vir die geel vet alleel?
Carriers | Draers = fr(Ww) = 2pq = 2(0.88)(0.12) = 0.21
0.21/[2(0.88)(0.12)+0.882] = 0.21
21% of the animals are carriers for the yellow fat allele | 21% van die diere is
draers vir geel vet alleel
Genetics | Genetika 214
3
May 2024
Tutoriaal 4 Kwantitatiewe- & Populasie Genetika
MEMO
1. A goldfish breeder is interested in two traits, tail-fin size (larger elaborate fins are deemed
more desirable) and resistance to fin rot (a fungal disease). His current breeding
population has an average fin length of 2 cm and on average fish take 7 days to recover
from infection. He runs a breeding experiment and selects fish with a mean tail fin length
of 4 cm and that have a mean recovery time of 3 days; the resulting offspring has a
mean tail fin length of 3.6 cm and a recovery time of 6.8 days. Which trait will be
improved by a continued selective breeding programme and why? Motivate with
appropriate calculations.
ʼn Goudvisteler stel belang in twee eienskappe, stertvingrootte (groter uitgebreide vinne
is meer gesog) en weerstand teen vinvrot ('n swamsiekte). Sy huidige broeipopulasie het
'n gemiddelde vinlengte van 2 cm en gemiddeld neem ʼn vis 7 dae om van infeksie te
herstel. Hy voer 'n teeleksperiment uit en kies vis met 'n gemiddelde vinlengte van 4 cm
en 'n gemiddelde hersteltyd van 3 dae; die gevolglike nageslag het ʼn gemiddelde
vinlengte van 3.6 cm en 6.8 dae herstel tyd. Watter eienskap sal verbeter word deur 'n
voortgesette selektiewe teelprogram en hoekom? Motiveer met toepaslike
berekeninge.
Tail-fin length | Stertvinlengte: Fin rot resistance | Vinvrotweerstand
h2 = (M2-M)/(M1-M) h2 = (M2-M)/(M1-M)
= (3.6-2)/ (4-2) = (6.8-7)/ (3-7)
= 0.8 = 0.05
Tail-fin-length seems to give a response to selection and this is supported by the realised
heritability estimate that is closer to one. On the contrary fin rot resistance has a realised
heritability closer to zero.
Stertvinlengte blyk ʼn respons op seleksie te gee en dit word ondersteun deur die
gerealiseerde oorerflikheidsberaming wat nader aan een is. Inteendeel, die
vinvrotweerstand het 'n gerealiseerde erflikheid van nader aan nul.
Genetics | Genetika 214
1
,Tutorial 4 Quantitative- & Population Genetics
May 2024
Tutoriaal 4 Kwantitatiewe- & Populasie Genetika
2. A researcher wants to investigate how much the environment influences stalk diameter
in ‘n wild strain of sugarcane. She grows the wild strain plants, with the cultivar N17 (highly
inbred) as an experimental control, in a field on the research farm. The phenotypic
variance for stalk diameter in the wild strain is 6.3 cm2, whilst the phenotypic variance for
N17 is only 2.1 cm2. What is the genotypic variance and the broad sense heritability in
the wild strain and what can be deduced about the influence of the environment on
phenotypic expression?
ʼn Navorser wil bepaal hoeveel die stingeldeursneë in ʼn wilde lyn suikerriet beïnvloed
word deur die omgewing. Sy groei plante van die wilde lyn met die van kultivar N17
(hoogs ingeteel), as ʼn eksperimentele kontrole, in die veld op die proefplaas. Die
fenotipiese variansie vir stingeldeursneë in die wilde lyn is 6.3 cm2, terwyl die fenotipiese
variansie vir N17 slegs 2.1 cm2 is. Wat is die genotipiese variansie en breë sin oorerflikheid
van die wilde lyn en wat kan afgelei word m.b.t. die omgewingsinvloed op fenotipiese
uitdrukking?
VP(wild) = 6.3 cm2
VP(inbred) = 2.1, in this line VG=0 (highly inbred), therefore VP = VE = 2.1
VP(wild) = VG + VE = VG + 2.1 = 6.3, therefore VG = 4.2 cm2
H2 = VG / VP = 4..3 = 0.67
Genetic differences between wild plants accounts for 67% of the observed variance in
phenotypic expression (broadsense heretibility). Thus the environment accounts for
33%.
Genetiese verskille tussen wilde plante verklaar 67% van die waargenome variansie in
fenotipiese uitdrukking (breë-sin oorerflikheid). Dus is die omgewing is verantwoordelik
vir 33%.
3. In a large flock of 5468 sheep, 76 animals have yellow fat, compared to the rest of the
members of the flock, which have white fat. It is assumed that this population is in HW
equilibrium for this autosomal locus. Furthermore, when sheep that are pure breeding for
white fat is mated to sheep that are pure breeding for yellow fat all of the resulting
offspring have white fat. When these F1’s are allowed to breed randomly, 25% of their
offspring have yellow fat.
In ʼn groot kudde van 5468 skape, het 76 diere geel vet, in vergelyking met die res van
die kudde wat wit vet het. Dit word aangeneem dat die populasie in HW ewewig is vir
hierdie outosomale lokus? Verder, wanneer suiwertelende wit vet skape geteel word
Genetics | Genetika 214
2
, Tutorial 4 Quantitative- & Population Genetics
May 2024
Tutoriaal 4 Kwantitatiewe- & Populasie Genetika
met suiwertelende geel vet skape, het al die nageslag wit vet. Wanneer hierdie F1’s
onderling lukraak teel, het 25% van hul nageslag geel vet.
a. What are the frequencies of the yellow- and white alleles in this population?
Wat is die frekwensie van die geel -en wit allele in die populasie?
Important: If both homozygous and heterozygous genotypic frequencies are given, the
formula {p = fr (p2) + ½ fr (2pq)} etc. must be used. You can only use the square root to
determine allele frequency if you cannot distinguish between the dominant
homozygotes and the heterozygotes.
Belangrik: Indien beide homosigotiese en heterosigotiese genotipiese frekwensies
gegee word moet jy {p = fr (p2) + ½ fr (2pq)} ens. gebruik om alleelfrekwensies te
bereken. Jy kan slegs die vierkantswortel gebruik om ‘n alleelfrekwensie te bepaal
indien daar nie onderskeid getref kan word tussen die dominante homosigote en die
heterosigote nie.
W = white fat | wit vet ; w = yellow fat | geel vet
Fr(WW) = p2
Fr(Ww) = 2pq
Fr(ww) = q2
q2 = = 0.014
q = √0.014
q = 0.12
p = 1-q = 0.88
b. What percentage of sheep with white fat are heterozygous carriers of the yellow fat
allele?
Watter persentasie van skape met wit vet is heterosigotiese draers vir die geel vet alleel?
Carriers | Draers = fr(Ww) = 2pq = 2(0.88)(0.12) = 0.21
0.21/[2(0.88)(0.12)+0.882] = 0.21
21% of the animals are carriers for the yellow fat allele | 21% van die diere is
draers vir geel vet alleel
Genetics | Genetika 214
3
