Question

In: Biology

In carrying out crosses with the nematode Caenorhabditis elegans, a geneticist hypothesizes that the allele for...

In carrying out crosses with the nematode Caenorhabditis elegans, a geneticist hypothesizes that the allele for dumpy body shape (dpy) is recessive to the allele for normal body (dpy+), and the allele for uncoordinated (unc) is recessive to the coordinated body movement allele (unc+). Moreover, she hypothesizes that these two genes are unlinked (i.e., they assort independently). In a mating between a normal bodied, coordinated worm and a dumpy, uncoordinated worm, all of the F1 worms were normal bodied and coordinated in their movements. Selfing these F1's (it should be noted here that C. elegans are capable of self- fertilization) yielded an F2 generation with the following phenotypes:

o normal body, coordinated 95

o normal body, uncoordinated 21

o dumpy body, coordinated 19

o dumpy body, uncoordinated 25

TOTAL 160

a. What is the expected ratio and number of each class of F2 worms for this cross?

b. Determine the Chi-square (X2) value for the observed data under the proposed hypothesis.

c. What is the probability value for this Chi-square value?

d. Do these data support the hypothesis that these loci are unlinked? Explain.

Expert Solution

Answer:

Based on the given information:

	dominant	recessive
normal body	dpy+	dpy	dumpy body
coordinated body movement	unc+	unc	uncoordinated body movement

F1: dpy+ unc+ x dpy unc

Progeny are all: dpy+ dpy unc+ unc

F2 Self cross: dpy+ dpy unc+ unc x dpy+ dpy unc+ unc

Punnett Square for F2 cross:

	dpy+ unc+	dpy+ unc	dpy unc+	dpy unc
dpy+ unc+	dpy+ dpy+ unc+ unc+	dpy+ dpy+ unc+ unc	dpy+ dpy unc+ unc+	dpy+ dpy unc+ unc
dpy+ unc	dpy+ dpy+ unc+ unc	dpy+ dpy+ unc unc	dpy+ dpy unc+ unc	dpy+ dpy unc unc
dpy unc+	dpy+ dpy unc+ unc+	dpy+ dpy unc+ unc	dpy dpy unc+ unc+	dpy dpy unc+ unc
dpy unc	dpy+ dpy unc+ unc	dpy+ dpy unc unc	dpy dpy unc+ unc	dpy dpy unc unc

Phenotype:

normal coordinated	normal coordinated	normal coordinated	normal coordinated
normal coordinated	normal, uncoordinated	normal coordinated	normal, uncoordinated
normal coordinated	normal coordinated	dumpy coordinated	dumpy coordinated
normal coordinated	normal, uncoordinated	dumpy coordinated	dumpy uncoordinated

Phenotype	Frequency
normal body, coordinated movement	9
normal body, uncoordinated movement	3
dumpy body, coordinated movement	3
dumpy body, uncoordinated movement	1

Phenotypic ratio 9:3:3:1

Chi Square Test:

Phenotype	Observed Frequency (O)	Expected Frequency ('E)	(O-E)	(O-E)^2/E
normal body, coordinated movement	95	90	5	0.278
normal body, uncoordinated movement	21	30	-9	2.700
dumpy body, coordinated movement	19	30	-11	4.033
dumpy body, uncoordinated movement	25	10	15	22.500
Total	160	160	Chi Square (sum)	29.511

Degree of freedom = number of categories - 1 = 4 - 1 = 3

P-Value corresponding to Chi square value of 29.511 with 3 degree of freedom is P-Value is < 0.00001.

The result is significant at a significance level of 0.05. Therefore Null Hypothesis is rejected. Based on the evidence, there is significant difference in observed frequency from expected frequency.

gladiator answered 2 years ago

In the nematode C. elegans, the vulva forms from a group of six vulval precursor cells...

In the nematode C. elegans, the vulva forms from a group of six vulval precursor cells (VPCs). These cells give rise to the three vulval lineages: 1º (central vulva), 2º (lateral vulva), and 3º (hypodermis). Experiments have determined that these cells constitute an “equivalence group”, meaning that they are each capable of forming all three vulval lineages. The crucial factor in their fate decision turns out to be their proximity to the anchor cell, which is a cell located in...

How do geneticist figure out where and when a gene is expressed? How does a geneticist...

How do geneticist figure out where and when a gene is expressed? How does a geneticist finds out what genes transcription are affected by a specific TF?

7. What proportion of the progeny of the following crosses, in which the different allele pairs...

7. What proportion of the progeny of the following crosses, in which the different allele pairs are segregating independently, will be homozygous? Aa Bb x Aa Bb Aa Bb CC x Aa Bb cc Aabbcc x AA bb CC

A recessive allele in mice results in abnormally long necks and sometimes death during embryogenesis. Crosses...

A recessive allele in mice results in abnormally long necks and sometimes death during embryogenesis. Crosses were made between pure breeding normal mice and pure-breeding long neck mic. All of the F1 mice had normal necks. Following an F1 x F1 cross, the F2 data below was observed for all mice born alive. 1044 Mice with normal necks 123 mice with long necks How many total mice were expected in the progeny? The answer is 348, how do we get...

Start by writing out your theoretical crosses for your fly traits. You should have four crosses...

Start by writing out your theoretical crosses for your fly traits. You should have four crosses for each reciprocal cross: •Mutant female x wild type male both autosomal •Mutant female x wild type male both X-linked •Mutant female x wild type male traitAautosomal, trait B X-linked •Mutant female x wild type male trait B autosomal, trait A X-linked •Mutant male x wild type female both autosomal •Mutant male x wild type female both X-linked •Mutant male x wild type female...

Start by writing out your theoretical crosses for your fly traits. You should have four crosses...

Start by writing out your theoretical crosses for your fly traits. You should have four crosses for each reciprocal cross: •Mutant female x wild type male both autosomal •Mutant female x wild type male both X-linked •Mutant female x wild type male traitAautosomal, trait B X-linked •Mutant female x wild type male trait B autosomal, trait A X-linked •Mutant male x wild type female both autosomal •Mutant male x wild type female both X-linked •Mutant male x wild type female...

Start by writing out your theoretical crosses for your fly traits. You should have four crosses...

Start by writing out your theoretical crosses for your fly traits. You should have four crosses for each reciprocal cross: •Mutant female x wild type male both autosomal •Mutant female x wild type male both X-linked •Mutant female x wild type male traitAautosomal, trait B X-linked •Mutant female x wild type male trait B autosomal, trait A X-linked •Mutant male x wild type female both autosomal •Mutant male x wild type female both X-linked •Mutant male x wild type female...

the act of carrying out the planned interventions

Anthropology: 1) If two parents are each carrying one copy of the allele for yellow eyes...

Anthropology: 1) If two parents are each carrying one copy of the allele for yellow eyes (Y) and for blue eyes (y), what are ALL the possible phenotypes of their children? (Assume that eye color is a simple, mendelian trait). a) Yellow Only b) Green Only c) yellow and blue only d) Yy e) YY, Yy, yy I picked yellow only but I guess I got that wrong 2) If a child is heterozygous for a particular trait, and the...

Explain the process of carrying out a behavioral study

Question

In carrying out crosses with the nematode Caenorhabditis elegans, a geneticist hypothesizes that the allele for...

Solutions

Expert Solution

Related Solutions

In the nematode C. elegans, the vulva forms from a group of six vulval precursor cells...

How do geneticist figure out where and when a gene is expressed? How does a geneticist...

7. What proportion of the progeny of the following crosses, in which the different allele pairs...

A recessive allele in mice results in abnormally long necks and sometimes death during embryogenesis. Crosses...

Start by writing out your theoretical crosses for your fly traits. You should have four crosses...

Start by writing out your theoretical crosses for your fly traits. You should have four crosses...

Start by writing out your theoretical crosses for your fly traits. You should have four crosses...

the act of carrying out the planned interventions

Anthropology: 1) If two parents are each carrying one copy of the allele for yellow eyes...

Explain the process of carrying out a behavioral study