Question

On the island of Notion, in the Factotum Archipelago, there lives a population of pencil bears....

On the island of Notion, in the Factotum Archipelago, there lives a population of pencil bears. A portion of these bears have silky fur, and as a result, are much better surfers. The texture of the fur is determined by a single autosomal locus with two alleles: S (trait allele) and s (wild type allele). In a recent study, researched collected the values in the following table:

Table 1. Counts of Pencil Bear fur style by genotype
Phenotype
Genotype Silky Fur Wild-type Fur
SS 90 10
Ss 100 50
ss 10 90

The S allele frequency in this population is 0.20, and the genotypes are in Hary-Weinberg proportions at the locus.  

If two heterozygous (Ss) bears mate, what proportion of their offspring who have wild-type fur will be have the ss genotype?

A.

0.54.

B. 0.50.

C. 0.57.

D. 1.00.

E. 0.04.

Reset Selection

1.For a pair of genes with alleles:

A (dominant)
a (recessive)

at the first locus; and

B (dominant)
b (recessive)

at the second locus,

that operate in a duplicate dominant epistatic manner, what proportion of offspring from a doubly-heterozygous mating are expected to show the recessive phenotype?

A. 9/16.

B. 1/16.
C. 0.5.
D. 1 (it's dominant).

E. 7/16.

7.

On the planet Kermit, in the Muppet population, there are two autosomal loci of interest. The first locus has alleles A and a, and the second locus has alleles B and b. The following phenotypes are associated with the alleles:
Locus 01
A = Yellow hair;
a = Orange hair.

Locus 02
B = Green skin;
b = Blue skin.

The A and B alleles are dominant for their respective loci.

Suppose a true breeding Yellow-haired, Green-skinned Muppet male mates with a true breeding Orange-haired, Blue-skinned Muppet. The couple produces multiple offspring, and a pair of the F1 Muppets intercross to produce 500 offspring. The counts for each pair of phenotypes are as follows:


Table 1. Counts of F2 phenotype pairs from F1 Muppet intercross  
Phenotype pair Count
Yellow hair, Green skin 282
Yellow hair, Blue skin 0
Orange hair, Green skin 93
Orange hair, Blue skin 125

The results presented in Table 1 suggest that:

A. Locus 01 and Locus 02 are on different chromosomes.

B. Hair color is independent of and individual's Locus 01 genotype for all Muppets.

C. Locus 01 and Locus 02 act independently in this population.

D. Skin color is independent of an individual's Locus 02 genotype for all Muppets.

E. Locus 01 and Locus 02 act epistatically in this population.

Reset Selection
3.Fluffernutter labradors are identical to Earth labradors in every way, except that they grow marshmallows instead of fur. The three possible types of marshmallow skin are:

a. Black (tastes like licorice);
b. Chocolate (tastes like chocolate);
c. Yellow (tastes like vanilla).

There are two genes that control for this phenotype, acting in a recessive epistatic manner. The two-locus genotypes that correspond to the phenotypes are:

A_B_ (codes for Licorice);
aaB_ (codes for Chocolate);
__bb (codes for Vanilla).

Note that the alleles at the two loci are:

A (dominant), a (recessive), (Locus 01)
B (dominant), b (recessive). (Locus 02)

If two doubly heterozygous Licorice Labs mate, what is the expected proportion of Licorice fluffernutter progeny?

A. 3/16.

B. 3/4.

C. 1/8.

D. 9/16.

E. 1/2.

4. OB01-Kenobi-ism in Jedis is controlled by a recessive gene (a). The other allele at the locus is A (Wild-type). Being homozygous for the a allele (genotype = aa) increases the force in a Jedi, whereas individuals who have one of the other two genotypes (Aa or AA) have standard level force values.

From mating between two non-OB01 carriers (both have Aa genotypes), five children are born and raised.

What is the probability that: at least one of the children has genotype aa? Hint: list the probabilities of the possible genotype frequencies among the offspring.

A. 0.763

B.

0.050



C. There is no solution.

D. 0.095

E. 0.237

5.

For the alien species below, matings are like humans in that the probability of transmitting either of the two alleles at a single locus is 0.50.

Here, we consider a single gene, MVE, with two alleles:

M = Individual prefers to bike-ride for exercise (dominant );

m = Individual prefers to walk for exercise (recessive).

All parental matings are true-breeding dominant crossed with true-breeding recessive.

In the population of Conurian Coupon collectors, a pair of parents mates and produces an F1 generation. The F1 generation inter-crosses to produce the following counts of F2 offspring:

Bike-riders: 73

Walkers: 27

What table most accurately reflects the goodness of fit design that will test whether the mode of the inheritance at this gene is autosomal dominant? That is, what table provides the most accurate observed and expected values, and the correct subsequent calculations?

A.
Phenotype Observed Expected O - E (O-E)^2 (O-E)^2/E
Bike-Riders 73 73 0 0 0
Walkers 27 27 0 0 0
Chi-square: 0
B.
Phenotype Observed Expected O - E (O-E)^2 (O-E)^2/E
Bike-Riders 73 25 48 2304 92.16
Walkers 27 75 -48 2304 30.72
Chi-square: 122.88
C.
Phenotype Observed Expected O - E (O-E)^2 (O-E)^2/E
Bike-Riders 73 75 -2 4 0.053
Walkers 27 25 2 4 0.16
Chi-square: 0.213
D.
Phenotype Observed Expected O - E (O-E)^2 (O-E)^2/E
Bike-Riders 73 50 23 529 10.58
Walkers 27 50 -23 529 10.58
Chi-square: 21.16
E.
Phenotype Observed Expected O - E (O-E)^2 (O-E)^2/E
Bike-Riders 73 99 -26 676 6.828
Walkers 27 1 26 676 676
Chi-square: 682.828


6.Which of the following statements is true regarding the inheritance of a human X-linked recessive trait if the mother is a carrier and the father is unaffected?

A. All sons will be affected.

B. None of the offspring will be affected.

C. On average, half of the daughters will be carriers and half of the sons will be affected.

D. On average, half of the daughters will be affected and half of the sons will be affected.

E. All daughters will be affected.

            

7.

If a marker locus with two alleles A and B and respective allele frequencies p and q is in Hardy-Weinberg Equilibrium, the population BB genotype frequency is:

A.

1-q2.

B.

q2.

C.

p2 + q2.

D. q.

E.

2pq.

Agapanthus is a dominant, homozygous lethal trait among Telorites from the planet Teleflora. Subjects with this condition love to send flowers. From a group of Telorites heterozygous at the Agapanthus locus (with two alleles: A = dominant (trait) allele, a = recessive (wild-type) allele), 88% of the Telorites have the Agapanthus phenotype, and 12% look wild-type.

Also, true-breeding wild-type Telorites (aa genotype) have 10% penetrance.

A Telorite with the Agapanthus phenotype mates with a true-breeding Telorite with the wild-type phenotype. One of the F1 children is randomly selected and is mated back to the true-breeding wild-type parent. Out of twenty F2 individuals, we only observe nineteen wild-type phenotype and one Agapanthus phenotype.

The genotype of the randomly-selected F1 subject must be:

A. aa.

B. Aa or aa.

C. AA.

D. We cannot say with certainty. We need to perform a Chi-square goodness of fit test.

E. Aa

0 0
Add a comment Improve this question Transcribed image text
Answer #1

Answer :

0.54

Explanation:

Ans 1: Cross bet ween Ss x Ss For uild type fun, we haue 100 4시50 气-( 100 400 600 100 0,025 t 0.166 + o.225 Thue, o4lb wie No

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