Tutorial 2
Tutoriaal 2
Mendelian Extensions
Mendeliese Uitbreidings 20 Feb. 2019
1.
A rabbit breeder mates a pure breeding doe with straight ears and short fur to a pure breeding buck with floppy ears and curly fur. The resulting kits all have straight ears with fluffy fur. When these kits (as adults) are allowed to interbreed, the following offspring are observed: ʼn Konynteler paar ʼn suiwertelende wyfie, met reguit ore en kort pels, met ʼn suiwertelende mannetjie met slap ore en ‘n krul pels. Al die nageslag het reguit ore met ʼn dons pels. Wanneer
hierdie nageslag toegelaat word om te interteel, word die volgende nageslag waargeneem:
Observed Phenotypic Count
Waargenome Fenotipiese Telling
75
Observed Phenotype
Waargenome Fenotipe
Straight ears, short fur
Reguit ore, kort pels
Straight ears, fluffy fur
Reguit ore, dons pels
Straight ears, curly fur
Reguit ore, krul pels
Floppy ears, short fur
Slap ore, kort pels
Floppy ears, fluffy fur
Slap ore, dons pels
Floppy ears, curly fur
Slap ore, krul pels
143
70
26
52
23 a.
Propose a hypothesis for the inheritance of these traits.
Stel ʼn hipotese voor vir die oorerwing van hierdie eienskappe.
Straight ears are dominant and Floppy ears recessive
Reguit ore is Dominant en Slap ore resessief
The fur trait is inherited in a codominant or incomplete dominant manner (depending an allelic function/action)
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Tutorial 2
Tutoriaal 2
Mendelian Extensions
Mendeliese Uitbreidings 20 Feb. 2019
Die pels eienskap word oorgeërf in ʼn kodominante of onvolledige dominante manier
(afhangend van die alleliese funksie/aksie) b.
What are the genotypes for the P1, the F1, and the F2 generations?
Wat is die genotipes vir die P, die F1, F2 generasies?
Ear trait | Oor eienskap: E/e
Coat trait | Pelseienskap: C/c
P1: EECC x eecc
F1: EeCc
F2:
Observed Phenotype
Waargenome Fenotipe
Genotype
Genotipe
Straight ears, short fur
Reguit ore, kort pels
Straight ears, fluffy fur
Reguit ore, dons pels
Straight ears, curly fur
Reguit ore, krul pels
Floppy ears, short fur
Slap ore, kort pels
Floppy ears, fluffy fur
Slap ore, dons pels
Floppy ears, curly fur
Slap ore, krul pels
E-CC
E-Cc
E-cc eeCC eeCc eecc c.
Test your hypothesis statistically.
Toets jou hipotese statisties.
Calculated expected ratios, do a chi-square analysis
Bereken die verwagte verhoudings en doen ‘n chi-kwadraat analise
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Tutorial 2
Tutoriaal 2
Fenotipes
Phenotypes
Straight ears, short fur
Reguit ore, kort pels
Straight ears, fluffy fur
Reguit ore, dons pels
Straight ears, curly fur
Reguit ore, krul pels
Floppy ears, short fur
Slap ore, kort pels
Floppy ears, fluffy fur
Slap ore, dons pels
Floppy ears, curly fur
Slap ore, krul pels
Totaal | Total
143
70
26
52
23
389
Mendelian Extensions
Mendeliese Uitbreidings
Observed
Waargeneem
75
Expected
Verwag d 2 /e
20 Feb. 2019
X 2 =
Df = 6-1 = 5
P > 0.05
Fail to reject H0. Observed deviation from the expected ratios is due to chance. The hypothesis for mode of inheritance holds.
Versuim, om die H0 te verwerp. Die waargenome afwyking van die verwagte verhoudings is a.g.v. kans. Die hipotese m.b.t. oorerwing volstaan. d.
A master’s student interested in identifying the genes responsible for these traits has narrowed it down to two possible candidate genes: Y and Z. Both these genes may have a role in synthesising structural proteins associated with cartilage and keratein. The Y gene allelic variants constitute a C to T base pair substitution in the nucleic acid sequence that creates a premature stop codon and the Z gene allelic variants constitute a G to A nonsynonymous substitution that alters the amino acid in the peptide chain. The student sequenced the DNA in each of the F2 animals and found the following: ʼn Meesterstudent is geïnteresseerd in die gene wat verantwoordelik is vir die eienskappe en
het sy kortlys van twee moontlike kandidaat gene geïdentifiseer: Y en Z. Beide die gene
mag moontlik ʼn rol speel in kraakbeen en keratien strukturele-proteïensintese. Die Y geen se alleliese variante blyk ʼn C tot T basispaar substitusie, in die nukleïensuur volgorde, te wees,
wat lei tot ʼn voortydige stop kodon. Die Z geen se alleliese variante is ‘n A tot G nie-
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Tutorial 2
Tutoriaal 2
Mendelian Extensions
Mendeliese Uitbreidings 20 Feb. 2019 sinonieme substitusie wat die aminosuur in die peptied ketting verander. Die student het die
DNS volgorde van elk van die F2 diere bepaal en het die volgende gevind:
Observed Phenotype
Waargenome
Fenotipe
Observed Phenotypic Count
Waargenome Fenotipiese
Telling
Observed DNA Sequence Genotype
(Y_Z) ratio
Waargenome DNS Volgorde
Straight ears, short fur
Reguit ore, kort pels
Straight ears, fluffy fur
Reguit ore, dons pels
Straight ears, curly fur
Reguit ore, krul pels
Floppy ears, short fur
Slap ore, kort pels
75
143
70
26
Genotipes (Y-Z) verhouding
1:2; C/C_A/A : C/T_A/A
1:2; C/C_A/G : C/T_A/G
1:2; C/C_G/G : C/T_G/G
All T/T_A/A
Floppy ears, fluffy fur
Slap ore, dons pels
Floppy ears, curly fur
Slap ore, krul pels
52
23
All T/T_A/G
All T/T_G/G
Which gene is responsible for which trait? Explain.
Watter geen is verantwoordelik vir watter eienskap? Verduidelik.
Ear trait is caused by Gene Y; and fur trait by gene Z (compare expected genotypic ratios with genotypic ratio of Sequence data)
Oor eienskap is a.g.v. Geen Y; en pels eienskap a.g.v. Z geen (vergely verwagte genotipe met genotipe vir die volgorde data) e.
Speculate on a possible biochemical mechanism that might be explained by the mode of inheritance and phenotypic observation.
Spekuleer oor ʼn moontlike biochemiese meganisme wat verklaar kan word deur die tipe
oorerwing en die fenotipiese waarnemings.
Ear trait: Dominant-recessive scenario, possibly a loss of function mutation
Oor eienskap: Dominant- resessiewe scenario, moontlik ʼn verlies van funksie mutasie
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Tutorial 2
Tutoriaal 2
Mendelian Extensions
Mendeliese Uitbreidings 20 Feb. 2019
Fur trait: If it is codominant, it could be due to the simultaneous expression of two fur fibre types; or if it is incomplete dominance it could be due to the expression of a “hybrid” fibre type, depending on mechanism of fur fibre production.
Pels eienskap: As dit kodominant is, kan dit a.g.v. die gelyktydige uitdrukking van twee pelsvel tipes wees; of as dit onvolledige dominansie is, a.g.v die uitdrukking van ʼn “hibried”
pelsvesel tipes, afhangend van hangend van die meganisme van pelsvesel produksie
2.
Flower colour in Proteas is determined at the Q locus where multiple alleles are known. The shape of Protea flowers is determined by the D locus. The alleles at both loci have the following effect:
Blomkleur by Proteas word bepaal deur die Q-lokus waar veelvuldige allele bekend is. Die vorm
van Protea blomme word bepaal deur die D-lokus. Die allele by beide loki het die volgende effek:
Bloedrooi
Wit
Pienk
Geel
Silwer
Colour
Kleur
Q q q p q y q s
Bloodred Enkel
White
Pink
Yellow
Silver
Langwerpig
Dubbel
Shape
Vorm
D d 1 d 2
Single
Long
Double
And you know that the dominance is as follows:
En jy weet dat die dominansie as volg is:
Q > q > q p > q y > q s & D > d 1 > d 2
Determine the genotypes of the P1 and F1 plants in the table below and predict the offspring
(phenotype + genotype) for the crosses between the different F1 plants as given.
Bepaal die genotipes van die P1 en F1 plante in die onderstaande tabel en voorspel die nageslag
(fenotipe + genotipe) vir die kruisings tussen die verskillende F1-plante soos gegee.
(NOTE: Not all possible F1 phenotypes are listed in the table)
(L.W. Nie alle moontlike F1 fenotipes word in die tabel aangedui nie)
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Tutorial 2
Tutoriaal 2
P1 phenotypes | fenotipes
silwer, enkel X wit, dubbel 
silver, single X white, double
wit, enkel X silwer, enkel 
white, single X silver, single
Mendelian Extensions
Mendeliese Uitbreidings
F
1
phenotypes | fenotipes
pienk, langwerpig
pink, long
X
geel, dubbel
yellow, double silwer, enkel X wit, dubbel silver, single X white, double q s q s Dd 1 X qq p d 2 d 2

 pienk, langwerpig
pink, long q p q s d 1 d 2

20 Feb. 2019
F2? wit, enkel X silwer, enkel  geel, dubbel
white, single X silver, single yellow, double qq y Dd 2 X q s q s Dd 2  q y q s d 2 d 2 gamete/s : q y d 2 ; q s d 2
F2
Fenotipe | Phenotype
Pienk, langwerpig
Pink, long
Pienk, dubbel
Pink, double
Geel, langwerpig
Yellow, long
Geel, dubbel
Yellow, double
Silwer, langwerpig
Silwer, long
Silwer, dubbel
Silwer, double
Genotipe | Genotype q p q y d 1 d 2 q p q s d 1 d 2 q p q y d 2 d 2 q p q s d 2 d 2 q y q s d 1 d 2 q q q y s s q q q s s s d
d
d
2
1
2 d d d
2
2
2
3.
Whilst trying to understand the inheritance of the yellow coat colour in mice, researchers mated two yellow heterozygous mice. A typical result was 56 yellow progeny to 31 wild-type. Use a chisquare test to determine if the outcome of this cross is consistent with the usual 3:1 ratio predicted by Mendelian inheritance for a dominant gene or could the colour gene be recessive lethal.
Om die oorerwing van geel pelskleur in muise te verstaan het navorsers twee geel heterosigotiese muise gekruis. ’n Tipiese resultaat was 56 geel nageslag en 31 wilde tipe nageslag. Gebruik die Chikwadraat toets om te bepaal of die uitkoms van hierdie kruising in lyn is met die gewone 3:1 ratio
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Tutorial 2
Tutoriaal 2
Mendelian Extensions
Mendeliese Uitbreidings 20 Feb. 2019 voorspel vir Mendeliese oorerwing met ’n dominante geen of kan die kleurgeen resessief letaal
wees?
Fenotipes
Phenotypes
Geel
Yellow
Wilde tipe
Wild type
Observed
Waargeneem
56
Expected
Verwag
3/4 (87) = 65.25
Chi square
Chi kwadraat
1.311
31 1/4 (87) = 21.75
(56 – 65.25) 2 /
65.25
(31 – 21.75) 2 /
21.75
3.934
Totaal / Total 87 544 5.25
Df = ?
P=?
Besluit: die verskil tussen die waarnemings en die verwagte ideale ratio is nie toevallig; ons verwerp die nulhipotese. ‘n 3:1 ratio kom nie voor nie.
Decision: the difference between the observed and expected ideal ratio’s are not due to chance; the null hypothesis is rejected. A 3:1 ratio does not occur.
Fenotipes
Phenotypes
Geel
Yellow
Wilde
Wild type
Totaal / Total tipe 31
87
Observed
Waargeneem
56
Expected
Verwag
2/3 (87) = 58
1/3 (87) = 29
544
(58 – 56) 2 / 58
(29 – 31) 2 / 29
Chi square
Chi kwadraat
0.069
0.138
0.207
Df = ?
P=?
Besluit: die verskil tussen die waarnemings en die verwagte ideale ratio is toevallig; ons verwerp nie die nulhipotese. ‘n 2:1 ratio kom voor.
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Tutorial 2
Tutoriaal 2
Mendelian Extensions
Mendeliese Uitbreidings 20 Feb. 2019
Decision: the difference between the observed and expected ideal ratio’s are due to chance; the null hypothesis is not rejected. A 2:1 ratio does occur.
4.
Mrs Mbeki and Mrs Zuma were roommates in the maternity ward, and both had daughters on the same day. After Mrs Mbeki took baby Jasmine home, she became convinced that the babies had been switched at the hospital. Blood tests were performed with the following results:
Mev. Mbeki en Mev. Zuma was kamermaats in die kraamsaal en het albei dogters gehad op dieselfde dag. Na Mev. Mbeki vir baba Jasmine huis toe geneem het, het sy oortuig geraak dat die
babas omgeruil is by die hospitaal. Bloedtoets resultate was as volg: o Mr and Mrs Mbeki were both type AB
Mnr. en Mev. Mbeki was albei tipe AB o Mr and Mrs Zuma were both type A
Mnr. en Mev. Zuma was albei tipe A o Jasmine Mbeki was type A and Victoria Zuma was type O
Jasmine Mbeki was tipe A en Victoria Zuma was tipe O
Were the babies switched?
Is die babas omgeruil?
A switch had not occurred. Explain
ʼn Omruiling nie plaasgevind nie. Verduidelik
5.
Coat colour in hamsters is determined by multiple alleles. If the order of dominance for the alleles are as follows, B (black) > b k (brown) > b d (cream) > b a (albino), determine the genotypes of each of the next parent pairs and indicate completely how the results were obtained:
Pelskleur in hamsters word deur veelvuldige allele bepaal. Indien die dominansie van die allele soos volg is, B (swart) > bk (bruin) > bd (room) > ba (albino) is, bepaal die genotipes van elkeen
van die volgende ouerpare en toon volledig aan hoe die resultate verkry is:
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Tutorial 2
Tutoriaal 2
Mendelian Extensions
Mendeliese Uitbreidings
Fenotipes van nageslag | Phenotypes of offspring
Ouers | Parents Swart | Black Bruin | Brown Room | Cream Albino
20 Feb. 2019
Swart x Swart
Black x Black
Bruin x Room
Brown x Cream
Swart x Albino
Black x Albino
Swart x Room
Black x Cream
22
0
13
14
0
24
0
12
0
11
12
0
7
12
0
0
Bb a X Bb a b k b a X b d b a
Bb d X b a b a
Bb k X b d b d / Bb k X b d b a
6.
In millet, colour is determined by three alleles at a single locus: Rp 1 (red), Rp 2 (purple), and rp
(green). Red is dominant over purple and green, and purple is dominant over green (Rp 1 >Rp 2 >rp).
Give the expected phenotypic and genotypic ratios of offspring produced by the following crosses:
In babalagras word kleur bepaal deur drie allele by ‘n enkele lokus: Rp 1 (rooi), Rp 2 (pers), en rp
(groen). Rooi is dominant oor pers en groen en pers is dominant oor groen (Rp 1 >Rp 2 >rp). Gee die verwagte fenotipiese en genotipiese ratios van die nageslag gevorm deur elkeen van die volgende kruisings: a.
Rp 1 /Rp 2 × Rp 1 /rp
¼ Rp 1 /Rp 1 (rooi | red), ¼ Rp 1 /rp (rooi | red), ¼ Rp 2 /Rp 1 (rooi | red), ¼ Rp 2 /rp (pers | purple),
Algehele fenotipiese ratio van ¾ rooi, ¼ pers
Overall phenotypic ratio of ¾ red, ¼ purple b.
Rp 1 /rp × Rp 2 /rp
¼ Rp 1 /Rp 2 (rooi | red), ¼ Rp 1 /rp (rooi | red), ¼ Rp 2 /rp (pers | purple), ¼ rp/rp (groen | green),
Algehele fenotipiese ratio van ½ rooi, ¼ pers, ¼ groen
Overall phenotypic ratio of ½ red, ¼ purple, ¼ green
Genetics | Genetika 214
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Tutorial 2
Tutoriaal 2 c.
Rp 1 /Rp 2 × Rp 1 /Rp 2
Mendelian Extensions
Mendeliese Uitbreidings 20 Feb. 2019
Hierdie is kruising tussen twee dubbele heterosigote so die verwagte fenotipiese ratio is ¾ rooi, ¼ pers
This cross involves a cross between two double heterozygotes, so the expected phenotypic ratio is ¾ red, ¼ purple d.
Rp 2 /rp × rp/rp
’n Kruising tussen heterosigoot en homosigotiese resessiewe individue (toetskruising).
Fenotipiese ratio van ½ pers, ½ groen
A cross of a heterozygote with a homozygous recessive (test cross). Phenotypic ratio is ½ purple, ½ green
7.
A family of six includes four children, each of whom has a different blood type: A, B, AB and O.
What are the genotypes of parents for this trait? ʼn Familie van ses sluit in vier kinders, elkeen met ʼn verskillende bloedgroep tipe: A, B, AB en O. Wat
is die ouers se genotipes vir hierdie eienskap?
Parents must be I A i and I B i
Ouers moet wees I A i en I B i
8.
Suppose two strains of Jimson Weeds are crossed and they produce a progeny in the following proportions. 360 spiny red, 724 spiny purple, 362 spiny white, 121 smooth red, 240 smooth purple,
120 smooth white. Explain the diverse phenotypes observed? What would be the genotype and phenotype of the above parental strain?
Veronderstel dat twee lyne Jimson onkruid gekruis word en hulle vorm nageslag in die volgende hoeveelhede: 360 rooi stekelrig, 724 pers stekelrig, 362 wit stekelrig, 121 rooi glad, 240 pers glad, 120 wit glad. Verduidelik hoekom soveel verskillende fenotipes waargeneem word? Wat sal die
genotipe en fenotipe wees van die ouerlike lyne?
The data can be summarised into an approximate ratio of 3:6:3:1:2:1. Parents SsRr x SsRr (both spiny purple).
Die data kan opgesom word as 3:6:3:1:2:1. Ouers SsRr x SsRr (albei stekelrig pers).
9.
The recessive lethal mutation "Manx" (Ma) in cats causes a tailless phenotype in the heterozygous condition. "Long hair" (lh) is a recessive mutation that determines hair length. Both are autosomal genes. Consider the cross (P) between a manx cat with normal, short hair; and a pure breeding
Genetics | Genetika 214
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Tutorial 2
Tutoriaal 2
Mendelian Extensions
Mendeliese Uitbreidings 20 Feb. 2019 cat with a normal tail, and long hair. Two F1 cats, that do not have the wild type phenotype, are mated to produce F2 offspring. Write the genotypes of these P and F1 cats and determine the phenotypic proportions for the F2 offspring.
Die resessiewe letale mutasie "Manx" (Ma) by katte veroorsaak stertloosheid in die heterosigotiese toestand. Lang hare (lh) is ʼn resessiewe mutasie wat haarlengte bepaal. Albei is outosomale gene.
Beskou die kruising (P) tussen ʼn manx kat met normale kort hare en ʼn suiwertelende kat met normale stert en lang hare. Twee F1 katte, wat nie die wilde fenotipe het nie, word gekruis om F2 nageslag te bekom. Skryf die genotipes van hierdie P en F1 katte en bepaal die fenotipiese
proporsies vir die F2 nageslag.
P1:
Ma+ +lh X ++ lhlh
F1:
Ma+ +lh manx (no tail); normal short hair |Manx (geen stert); normale kort hare
Ma+ lhlh manx (no tail); long hair |Manx (geen stert); lang hare
++ +lh normal tail; normal short hair |normale stert; normale kort hare
++ lhlh normal tail; long hair |normale stert; lang hare genotypes F
1
cats | genotipes F
1
katte: Ma+ lhlh x Ma+ lhlh
F
2
-generasie / generation:
Ma lh
MaMa lhlh
+ lh
Ma+ lhlh
Ma lh
+ lh Ma + lhlh ++ lhlh
2/3 manx (no tail); long hair |manx (geen stert); lang hare
1/3 normal tail; long hair | normale stert; lang hare
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