Question

1. People wit h the sex-linked genetic disease, hemophilia, suffer from excessive bleeding because their blood will not clot. Tom, Mary, and their 4 daughters do not exhibit symptoms of hemophilia. However, their son exhibits symptoms of hemophilia because: A) Tom is heterozygous B) Tom is homozygous C Mary is homozygous D) Mary is heterozygous E All of the above are equally likely What is the risk of having a child affected by disease with an autosomal recessive inheritance both the mother and father are carriers for the disease? A) 0% B) 25% 2. H n -. C) 5096 X- -E) 10096メ 3. Which of the following observations would support the theory of maternal inheritance for the spunky phenotype? A) Spunky female x wild-type male progeny all spunky B) Wild-type female x spunky male progeny all spunky C) Wild-type female x spunky male progeny spunky, % wild-type D) Spunky female x wild-type maleprogeny spunky, x wild-type E) Spunky female x wild-type maleprogeny all wild-type Which cross must produce all green, smooth peas if green (G) is dominant over yellow (g) and smooth (S) is dominant over wrinkled (s)? A) GgSs x GGSs B) GgSS x ggSS C) Ggss x GGSS D) GgSs x GgSs E) None of the above 4. The tall allele is dominant to the short allele. True breeding tall plants were crossed to true breeding short plants. The F1 plants were self-crossed to produce F2 progeny. What are the phenotypes of the F1 and F2 progeny? A) All F1 and 1/4 of the F2 are short B) All F1 are short and 1/4 of the F2 are tal C) All F1 and 3/4 of the F2 are tall 5. TS D) All F1 are tall and 3/4 of the F2 plants are short E) None of the above

Answer 1

1. Ans. Is (D) Mary is heterozygous.

Explanation:Haemophilia is a sex linked recessive genetic disease.The recessive allele is linked toX chromosome.As males contain one X chromosome and one Y chromosome if allele for haemophilia is present on X chromosome of male, person will be haemophilic.Since female contains two X chromosomes and allele for haemophilia is recessive if one X chromosome contains haemophilic allele , female will be carrier and wont show any symptoms.In the given case, as Tom and her sons are not having allele for haemophilia on their X chromosomes, they didn’t show any symptoms.As Mary being female containing two X chromosomes, even though containing allele for haemophilia on one X chromosome since other X does not contain allele and as haemophilic allele is recessive, Mary also didn’t show any symptoms.It is given that Mary’s son exhibits symptoms of haemophilia he should have got that allele from mother as from father son inherits only Y chromosome and X chromosome always be inherited from mother .As son is showing symptoms means he got allele from Mary who is not showing any symptoms Since son inherits X chromosome from mother he inherited X chromosome containing haemophilic allele from mother and Mary not showing symptoms of haemophilic indicates that Mary is heterozygous , if she is homozygous she should have shown symptoms of haemophilia.

2. (B) 25%

Explanation:as it is mentioned the disease is autosomal recessive, it is not sex linked and alleles arepresent on autosomes.As it is recessive and both the parents are carriers but not affected by disease probability of gamete carrying from both the parents is ½ (ie., 50% of gametes from mother and 50% of gametes from father can carry this allele and risk (probability) of having child affected by disease (as the disease is recessive, both gametes should carry allele) is ½ x ½ = ¼ which is 25%.

3. (A) Spunky female x wild type male   -------à progeny all spunky.

Explanation:As it is clearly mentioned as maternal inheritance the genes must be present in the cytoplasm of the female as genetic contribution will always be equal from both the parents andcytoplasmic inheritance will be only from female (reason for large size of female gamete compared to that of male gamete) when the female is spunky all progeny is spunky as cytoplasm from female is inherited to all the progeny. Whereas in other cases wild progeny is also seen which is not possible if female is spunky if inheritance is maternal.

4. (A) GgSs X   GGSS

Explanation:As Green and smooth are dominant over yellow and wrinkled , the progeny to be green and smooth if one G allele and one S allele is present in genotype.As all progeny are green and smooth atleast one G and one S should be present in their genotype which is only possible in the above combination as all gametes from parent (GGSS) will contain G and S which is not possible in other cases.

5.( C ) All F1 and 3/4^th F2 are tall.

Explanation:It is mentioned that Tall is dominant over short and both the parents are true breeding type that is one parent is pure tall (means homozygous genotype of tall TT) and other is pure dwarf (means homozygous genotype of dwarf tt) all F1 progeny due to presence of T in their genotype will be tall (as all gametes from tall parent contain T allele) and when selfing of F1 is done the progeny will be in the ration of 3:1 (Tall : Dwarf) as the probability of getting allele (either T or t) is ½ in each gamete due to which 1/4^th will be pupre tall and ½ will be heterozygous tall and 1/4^th will be dwarf and phenotypically both pure and heterozygous tall will be tall this option is correct.