Question

4.Using the in data table 1 for the F1 generation of the homozygous crosses identify, if any, genes that appear to be sex linked. Describe the reasoning you used to come to this conclusion. 5.Using the data in table 2 for the F1 generation of the heterozygote homozygote crosses identify, if any, genes that appear to be linked together. Describe the reasoning you used to come to this conclusion. 6. Using the data in table 3 determine the likely wild type for each characteristic.Explain the rationale for how you identified the wild type for each trait.  

table 1. Phenotype counts for homozygous cross

Initial Cross

Phenotypes

SSPP x sspp

236SP♂ 264SP♀

SSKK x sskk

249SK♂ 251SK♀

SSWW x ssww

131SW♂ 122Sw♂ 247SW♀

SSFF x ssff

239SF♂ 261 SF♀

SSEE x ssee

257SE♂ 243SE♀

PPKK x ppkk

255PK♂ 245PK♀

PPWW x ppww

119PW♂ 122Pw♂ 259PW♀

PPFF x ppff

247PF♂ 253PF♀

PPEE x ppee

252PE♂ 248PE♀

KKWW x kkww

125KW♂ 127Kw♂ 248KW♀

KKFF x kkff

259KF♂ 241KF♀

KKEE x kkee

251KE♂ 249KE♀

WWFF x wwff

120WF♂ 119wF♂ 261WF♀

WWEE x wwee

125WE♂ 125wE♂ 250WE♀

FFEE x ffee

249FE♂ 251FE♀

Table 2. Phenotype counts for heterozygous homozygous cross.

Initial Cross

Phenotypes

SSPP x sspp

263SP 271Sp 244sP 222sp

SSKK x sskk

253SK 265Sk 239sK 222sk

SSWW x ssww

263SW 271Sw 244sW 222sw

SSFF x ssff

263SF 271Sf 244sF 222sf

SSEE x ssee

263SE 271Se 244sE 222se

PPKK x ppkk

263PK 271Pk 244pK 222pk

PPWW x ppww

263PW 271Pw 244pW 222pw

PPFF x ppff

263PF 271Pf 244pF 222pf

PPEE x ppee

263PE 271Pe 244pE 222pe

KKWW x kkww

263KW 271Kw 244kW 222kw

KKFF x kkff

429KF 62Kf 69kF 440kf

KKEE x kkee

263KE 271Ke 244kE 222ke

WWFF x wwff

263WF 271Wf 244wF 222wf

WWEE x wwee

263WE 271We 244wE 222we

FFEE x ffee

263FE 271Fe 244fE 222fe

Table 3. Phenotypic frequencies of characteristics in wild

Characteristic

Allele

Frequency

Allele

Frequency

Eye Stalk

Stalk(S)

89%

No stalk(s)

11%

Fur Pattern

Spotted(P)

7%

Striped(p)

93%

Spines

Spined(K)

75%

No spines(k)

25%

Wings

No wings(W)

97%

wings(w)

3%

Tail

Fluffy Tail(F)

12%

Leathery tail(f)

88%

Eye Color

Blue(E)

57%

Orange(e)

43%

Answer 1

Since we only answer 1 question in case of multiple questions, we’ll answer the first question as the exact one wasn’t specified. Please resubmit the question separately and specify the question you'd like answered.

Answer 4)

In given table 1the F1 homozygous crosses two traits. For two given homozygous trait (let us assume AABB * aabb), the F1 phenotype would be 1:0 (all would be AaBb). For a sex-linked trait, i.e if the allele for the trait is present in the X chromosome, the F1 will show a heterozygous female (XAXa) and a hemizygous male (XAY). So the ratio of males is to females must be 1:1. Now in the given table, we see that when the trait for wings ( W/w) is considered then the ratio for the males shows a difference than the other traits. The males are being divided into homozygous (W) and heterozygous (w). This indicates the trait for wings may be sex-linked.