data for first two pages: Monohybrid Cross: Purple hypocotyl - 942 Green hypocot

Question

data for first two pages:

Monohybrid Cross:
Purple hypocotyl - 942
Green hypocotyl - 406

Dihybrid Cross:
Purple hypocotyl, green cotyledons - 687
Purple hypocotyl, yellow cotyledons - 255
Green hypocotyl, green cotyledons - 310
Green hypocotyl, yellow cotyledons - 96

In Lab #6 you did a Chi-Square (X2) analysis of the monohybrid and dihybrid crosses of fast plants using your Lab Section Data. You will find it necessary to consult information within Lab #6 (still viewable on Blackboard) to complete this assignment. For the first part of this homework you will do a X2 analysis of the monohybrid and dihybrid crosses using data pooled from all of the lab sections (Entire Course Data). This data will be posted as an announcement on Blackboard and distributed as an email sent to all students. A. MONOHYBRID CROSS ANALYSIS Entire Course Data - Number with Purple hypocotyls Number with Green hypocotyls Total Number of seedlings 1. What possible genotype(s) could plants with Purple hypocotyls have? 2. What possible genotype(s) could plants with Green hypocotyls have? 3. Perform the calculations below to fill-in the X2 table: Phenotype: Purple Green Expected Value (e) Observed Value (o) Deviation (d) Deviation Squared (d2) X2 Value add all (d2+ e)

Explanation / Answer

A. Monohybrid cross analyses

Number of purple cotyledons= 942

Number of green cotyledons= 406

Total number of seedlings= 1348

1. what possible genotypes could plants with purple hypocotyls have?

Answer: PP or Pp

2. what possible genotypes could plants with green hypocotyls have?

answer:gg

3. Purple: expected value: 1348/4 *3= 1011

observed value: 942

observed -expected= 1011-942=69

(O-E)2= 4761

chi square: (O-E)2/E= 4761/1011= 4.709

Green

expected value: 1348/4*1= 337

Observed value: 406

(O-E)2= (406-337)2= 4761

chi square= (O-E)2/E= 4761/337= 14.12

4. Degree of freedom= Number of phenotypes -1

2-1=1

Chi square value for df=1 (0.05)= 3.84

5. chi square values obtained from calculations above are greater than table value so 3:1 hypothesis does not fit in.

B. Dihybrid cross analysis=

Dihybrid Cross:  

Observed value(O) Expected value(E) (O-E) (O-E)2 (O-E)2/E
Purple hypocotyl, green cotyledons - 687 758.25 71.25 5076.56 6.69
Purple hypocotyl, yellow cotyledons - 255 252.75 2.25 5.0625 0.02
Green hypocotyl, green cotyledons - 310 252.75 57.25 3277.56 12.96
Green hypocotyl, yellow cotyledons - 96 84.25 11.75 138.06 1.63

chi square value= sum of (O-E)2/E= 21.3

2. Degree of freedom= 4-1=3

chi square critical value for df 3= 7.82

3. 9:3:3:1 ratio cannot be accepted since chi square value is way higher than critical value

c. Genetics Problems

Problem 1. Monohybrid crosses

a) A homozygous hairy plant TT is pollinated by b) homozygous hairless tt plant

c) T allele is carried by egg d) t allele is carried by sperm e) genotype of F1 generation is Tt

f) phenotype will be hairy plant since hairy trait is dominant over hairless

g) If Tt is self pollinated phenotype of F2 will be both hairy and hairless

h) alleles carried by eggs of parents of F2= T and t

i) alleles carried by sperm of parent of F2= T and t

j) phenotypic ratio= 3 (hairy) and 1(hairless)= 3:1

k) genotypic ratio= 1 (homozygous hairy): 2(heterzygous hairy): 1(homozygous hairless)= 1:2:1

Problem 2

A blue flower plant is pollinated by ed flower plant half progeny are blue and half are red

Genotype of parents: blue (Ff) red (ff)

genotypes of progeny will be either Ff or ff

Problem 3 Dihybrid cross

Round seed with hairy plant is crossed with oblong seed with hairless plant

TTSS X ttss

a) alleles carried by eggs TS

b) alleles carried by sperm ts

c) TTSS X ttss

ts ts

TS TtSs TtSs

TS TtSs TtSs

d) genotype of F1: TtSs

e) phenotype of F1 hairy plant with round seed

f) TtSs X TtSs

TS Ts tS ts

TS TTSS TTSs TtSS TtSs

Ts TtSs TTss TtSs Ttss

tS TtSS TtSs ttSS ttSs

ts TtSs Ttss ttSs ttss

g) phenotype of F2 generation; round hairy, round hairless, oblong hairy and oblong hairless

h) phenotypic ratio 9:3:3;1

i) genotypic ratio 1:2:2:1:4:1:2:2:1

j) TTSS, TtSS, TtSs

Problem 4

Tall plant (TT)

Dwarf plant (tt)

TT X tt= Tt (intermediate height)

Tt X Tt

Progeny will be TT, Tt, Tt and tt so phenotypes will be 1 tall 1 dwarf and 2 intermediate height and ratio will be 1:2:1

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