Determine the Chi Suquare for this table, remember thet yellow mutant is a sex-lincked gene. Show...

Determine the Chi Suquare for this table, remember thet yellow mutant is a sex-lincked gene. Show all the work.

Cross 1: Dumpy Males x Yellow Females
	Males	Females	Total
Wild Type	23	27	50
Mutant A (Yellow )	14	5	19
Mutant B (Dumpy)	6	7	13
Double Mutant AB	4	2	6
Total	47	41

Expert Solution

Cross 1: Dumpy Males x Yellow Females
	Males	Females	Total
Wild Type	23	27	50
Mutant A (Yellow )	14	5	19
Mutant B (Dumpy)	6	7	13
Double Mutant AB	4	2	6
Total	47	41	88 (Grand Total)

Null hypothesis = The genes for color & height are assorted independently.

Phenotype	Number Observed	Number Expected
Wild type male	23	(50 x 47)/88 = 26.70
Wild type female	27	(50 x 41)/88 = 23.30
Mutant A (Yellow) male	14	(19 x 47)/88 = 10.15
Mutant A (Yellow) female	5	(19 x 41)/88 = 8.85
Mutant B (Dumpy) male	6	(13 x 47)/88 = 6.94
Mutant B (Dumpy) female	7	(13 x 41)/88 = 6.06
Double mutant AB male	4	(6 x 47)/88 = 3.20
Double mutant AB female	2	(6 x 41)/88 = 2.80

² = (Observed - Expected)²/Expected

= (23-26.70)²/26.70 + (27-23.30)²/23.30 + (14-10.15)²/10.15 + (5-8.85)²/8.85 + (6-6.94)²/6.94 + (7-6.06)²/6.06 + (4-3.20)²/3.20 + (2-2.80)²/2.80

= 0.51 + 0.59 + 1.46 + 1.67 + 0.13 + 0.15 + 0.2 + 0.23

= 4.94

Degree of freedom (df) = (number of rows – 1) x (number of columns – 1) = (4-1) x (2-1) = 3 x 1 = 3

We find that probability from ² distribution table is 0.1 < P < 0.5.

The probability associated with the calculated chi-square value is between 0.10 and 0.50, indicating a high probability that the difference between observed and expected values is due to chance.

gladiator answered 2 years ago

The following sequences show the wildtype and mutant alleles of part of a gene. What is...

The following sequences show the wildtype and mutant alleles of part of a gene. What is the name of this type of mutation? Wildtype DNA: 5’ AAC – AGC – CTG – CGT – ACG – GCT – CTC 3’ Wildtype protein: Asn – Ser – Leu – Arg – Thr – Ala – Leu Mutant DNA: 5’ AAC – AGC – CTG – CTT – ACG – GCT – CTC 3’ Mutant protein: Asn – Ser –...

Why do yellow gene mutations impact body color in the fly? Do the two mutant alleles...

Why do yellow gene mutations impact body color in the fly? Do the two mutant alleles from a female yellow fly need to be the same mutation (same allele)?

In chickens, there is a mutant gene called "frizzle" that results in weak, stringy, and easily...

In chickens, there is a mutant gene called "frizzle" that results in weak, stringy, and easily broken feathers. When a frightfully frizzled fowl is bred to a normal chicken, the offspring are mildly frizzled. If one breeds two mildly frizzled chickens to each other, the offspring have the phenotypic ratio of normal: 2 mildly frizzled: 1 frightfully frizzled. Indicate the parent phenotypes and genotypes, and draw a punnett square for this cross.

Cystic fibrosis is caused by a mutation in a single gene. Homozygotes for this mutant recessive...

Cystic fibrosis is caused by a mutation in a single gene. Homozygotes for this mutant recessive allele (ff) have cystic fibrosis. Within a population, 2% of people have cystic fibrosis. Assuming Hardy-Weinberg equilibrium, the frequency of heterozygotes in the population is 0.24. If inbreeding was common in this population, what would happen to the percentage of people who have cystic fibrosis over time?

Question text In drosophila (fruit flies): The yellow gene is on the X-chromosome, the gene product...

Question text In drosophila (fruit flies): The yellow gene is on the X-chromosome, the gene product plays a role in body colour. Wild type flies are a light brown/tan in colour, homozygote yellow mutants have a more yellow shaded body colour. We'll use y+ to represent the wild type allele, and y to represent the mutant allele. y+ is dominant to y. The vestigial gene is on chromosome 2. Homozygote vestigial mutants have very short wings, and cannot fly. We'll...

The normal (wildtype) size of the brat gene is 1500 base pairs (bp). In the mutant...

The normal (wildtype) size of the brat gene is 1500 base pairs (bp). In the mutant variant you identified, there is a deletion of 500 bps in the last exon of the gene. How would you recognize the two versions based on size? List what fragment sizes you would expect to see on a gel.

DNA sequence of wild type gene A and a mutant is shown below. 1. Transcribe and...

DNA sequence of wild type gene A and a mutant is shown below. 1. Transcribe and translate the wild type and mutant proteins. 2. Classify the type(s) of mutation(s) in gene A the mutant has. 3. Design a primer pair to generate a PCR fragment of any size from the wild type sequence. Write the primer pair and mark the 5’ and 3’ of each primer sequence. wild-type gene 5’ TAG|ACC|ATG|CCA|GTA|AAT|TTA|CGA|TGA|CA 3’ 3’ ATC|TGG|TAC|GGT|CAT|TTA|AAT|GCT|ACT|GT 5’ mutant 2 5’ TAG|ACC|ATG|CCA|GTA|AAT|ATG|TTA|CGA|TGA|CA 3’ 3’ ATC|GGG|TAC|GGT|CAT|TTA|TAC|AAT|GCT|ACT|GT...

Now you want to study the protein product of this mutant brat gene, and you want...

Now you want to study the protein product of this mutant brat gene, and you want to introduce this human brat gene into a bacterial expression vector (one that can be expressed to make a protein in bacterial cells) in hopes of producing a large amount of the human mutant Brat protein. Should you use the genomic DNA or the cDNA version of this gene? Explain your reasoning.

If you introduced a strongly expressed gene for a mutant enzyme that constitutively exhibited adenylate cyclase...

If you introduced a strongly expressed gene for a mutant enzyme that constitutively exhibited adenylate cyclase activity, what would you predict to be the relative level (high/normal/low) of β-galactosidase activity in the following growth conditions? Explain each answer. Presence of both glucose and lactose (3 points) Absence of glucose, presence of lactose (3 points) Absence of glucose and absence of lactose (3 points)

Among the alleles for the gene responsible for cystic fibrosis that have been identified, one mutant...

Among the alleles for the gene responsible for cystic fibrosis that have been identified, one mutant has been identified in which three amino acids in the middle of the protein sequence are changed. What change in the DNA sequence of the gene could cause such a mutation? (Choose the best answer): Select one: a. a single base pair deletion b. insertion of three adjacent base pairs c. a single base pair substitution d. a single base pair insertion and a single...

Question