OB01-Kenobi-ism in Jedis is controlled by a recessive gene (a). The other allele
ID: 73061 • Letter: O
Question
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
2.On the planet LongLiveRock, there is a recessive condition known as WhoAreYou. People with this trait have an uncontrollable perform windmill guitar swings. The two alleles at the trait locus are labeled R (trait) and r (wild type). If the probability of transmitting an R allele is 0.76, then what is the probability that any child born to carrier parents will be affected (i.e., will perform windmill guitar swings)?
A. 0.76 B. 0.58 C. 0.62 D. 0.42 E. 0.24
3.Among Urepian Rock-things, skin texture is determined by a single autosomal gene. The wild type allele of this gene (th+) codes for hard, scale-like skin, whereas a mutant allele (th) codes for soft, smooth skin. When a Rock-thing with hard skin is crossed with a Rock-thing whose phenotype is soft, smooth skin, only smooth, soft-skinned offspring are ever observed. One possible answer is that: (Hint: For each answer, see if the offspring phenotypes are consistent the parental matings).
A. the parental Rock-thing with the soft, smooth skin is homozygous at the gene and the th allele is dominant.
B. the parental Rock-thing with the hard skin is heterozygous for the gene and the th+ allele is dominant.
C. the parental Rock-thing with the hard skin is homozygous at the gene and the th+ allele is dominant.
D. the parental Rock-thing with the soft, smooth skin is heterozygous at the gene and the th+ allele is dominant.
E. neither allele is dominant
4. There is a population of alien dogs, species Dogus Cute-ee-us, that live on the planet Woof. Their genomes and biology are amazingly identical to the genomes and biology of earth dogs. Consider a di-hybrid cross, where the first locus has alleles A and a, and second locus has alleles B and b. Each additional copy of either an A or B allele in the two-locus genotype increases the friendliness of the dog. Friendliness is measured by the Wilcoxon-Canus Amicus Diagnostic Scale (WCADS). There are five categories: 0 = Unfriendly/Loner (although not aggressive); 1 = Somewhat friendly/shy; 2 = Friendly; 3 = Very friendly; 4 = Maximally friendly. For example, an alien dog with a two-locus genotype of Aabb would have a Somewhat friendly/shy behavior, since there is one copy of A and zero copies of B, adding to a score of 1 for the WCADS. Similarly, an alien dog with a two-locus genotype of AaBB has a Very friendly behavior (one copy of A + two copies of B = three copies total of A or B). In the following mating: AaBb x AaBb, what is the expected ratio of Maximally friendly:Very friendly:Friendly:Somewhat friendly:Unfriendly alien dogs in the offspring, assuming the parents produce a very large litter?
A. 1:2:3:2:1
B. Question cannot be answered with the information given.
C. 1:4:6:4:1
D. 9:3:3:1
E. 0:1:1:1:0
5.For an X-linked recessive trait, if a female fly who is a carrier for the trait-allele is crossed with a male who does not show the trait, what proportion of the offspring will show the trait?
A. 0.75 B. 0.25 C. 1 D. 0.5 E. 0
6.When Freddy the fly, who has big eyes (the mutant phenotype), mates with Fredonia the fly, who has small eyes, their offspring appear in the following proportions:
Table 1. Offspring phenotypes with proportions :Phenotype Proportion Big-eyes, female 25% Small-eyes, female 25% Big-eyes, male 25% Small-eyes, male 25%
Note that the gene for eye size is autosomal. Fredonia is homozygous at the eye size locus. The most likely explanation for observing these proportions is:
A. Fredonia doesn't like Freddy.
B. The gene coding for big eye size is autosomal recessive trait.
C. The gene coding for big eye size is X-linked recessive.
D. The gene for big eye size is autosomal dominant.
E. The genes coding for big eye size and gender appear on the same chromosome.
7.Given genes A, B, and C with phenotypes: Gene A Phenotype: Triceratops leg curvature; Gene B Phenotype: Farfeggnuggen wing vein; Gene C Phenotype: Muppeteritus antennae. And given the following table of phenotype counts (where "+" refers to wild-type phenotype):
ABC 326
AB+ 6
A+C 26
A++ 124
+BC 121
+B+ 32
++C 4
+++ 314
which gene is in the middle?
A. B.
B. There is no middle gene. A is on a different chromosome than B and C
C. C
D. A.
E. There is no middle gene. A is linked to B, but not to C.
Explanation / Answer
1. We have to do the following cross: Aa x Aa which will give us the result AA : Aa : aa in the ratio 1:2:1.
therefore the probability of each genotype is 1/4, 2/4, 1/4.
question asks us to tell the probability of having atleast one aa genotype which could be obtained by doing:
1 - probability of all dominant
now probability of all dominant = 3/4 and if all 5 are dominant, then probability becomes (3/4) ^5 which is 0.23
so 1- 0.23 = 0.77 or 0.763 hence option A.
2. In this question frequency of R = 0.76 so the frequency of r = 0.24 . In a cross of Rr x Rr , there would be both Rr and RR who would not perform guitar swings.
probability of RR = 0.76 x 0.76 = 0.5776
probability of Rr = 0.76 x 0.24 = 0.1824
probability rr = 1 - probability of RR and Rr
therefore it is 1 - 0.76 = 0.24 therefore it is option E.
3. here the gene th+ codes for hard skin and th codes for soft skin.
when we cross a hard skin with soft skin organism, we obtain all soft skin rock things, this shows that th allele is dominant to th+
the parental cross could be shown like this:
thth x th+th+ this will give all the genotypes as th th+
which shows option A is correct.
4. since the cross says we cross AaBb x AaBb
which will give the results: AABB = 1/16 = maximally friendly , aabb = 1/16 = unfriendly, very friendly ones will have 3 capital letters which are 4 in number , the friendly ones will have a score of 2 which means two capital letters which are 6 in number. somewhat friendly ones will have only 1 capital letter which are 4 in no. so the ratio would be Maximally friendly:Very friendly:Friendly:Somewhat friendly:Unfriendly = 1:4:6:4:1
therefore the correct option is C.
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