Present value calculations allow one to determine A. the return to an uncertain asset. B. the present-day costs and/or benefits of various options. C. the utility value of a particular option. D. the social cost of financial calculations. E. the real wage. Solution B. the present-day costs and/or benefits of various options. Present value calculations involve discounting future costs and/or benefits of a project so that they can be compared with other available alternatives..

1. Problem 13.32 Copy For the following crosses, determine as accurately as possible the
genotypes of each parent, the parent in whom nondisjunction occurs, and whether nondisjunction
takes place in the first or second meiotic division. Both color blindness and hemophilia, a blood-
clotting disorder, are X-linked recessive traits. In each case, assume the parents have normal
karyotypes (see the table given below) Table Human Aneuploidiesand Frequencies at Birth.
Aneuploidy Syndrome Frequency at Birth Syndrome Characteristics Autosomal Aneuploidy
Trisomy 13 Patau syndrome 1 in 15,000 Mental retardation and developmental delay, possible
deafness, major organ abnormalities, early death Trisomy 18 Edward syndrome lin 8000 Mental
retardation and developmental delay. skull and facial abnormalities, early death Trisomy 21
Down syndrome Mental retardation and developmental delay characteristic facial abnormalities,
short stature, variable life span Sex-Chromosome Aneuploidy Klinefelter syndrome 1 in 1000
(males) Variable secondary sexual characteristics, infertility, frequent breast swelling: no impact
on mental capacity 47, XXY 47, XYY Jacob syndrome lin 1000 (males) Tall stature common:
possible reduction but not loss of fertility: no impact on mental capacity Triple x syndrome in
1000 (females) Tall stature common: possible reduction of fertility; menstrual imegularity: no
impact on mental capacity Tumer syndrome in 5000 (females) No secondary sexual
characteristics infertility, short stature; webbed neck common; no impact on mental capacity Part
A Aman and a woman who each have wild-type phenotypes have a son with Klinefelter
syndrome (XXY who has hemophilia The man is XH Y the woman is X XH. the child is xixi y,
and nondisjunction occurred in the woman during meiosis II The man is XH the woman is X xii.
the child is xhix l y, and nondisjunction occurred in the woman during meiosis I O The man is
XH y, the woman is xi the child is x x Y, and nondisjunction occurred in the woman during
meiosis II O The man is xi the woman is x xii the child is x Y, and nondisjunction occurred in
the woman during meiosis I Submit My Answers Give Up
Solution
PART A
A man and a woman who each have wild type phenotype have a son with klinefelter syndrome
(XXY) who has hemophilia
Answer : option A
genotype of the mother XHXh
genotype of the father XHY
nondisjunction occurED in the mother
PART B
A man who is color blind and a woman who has a wild-type phenotype have a sonwith Jacob

2. syndrome (XYY) who has hemophilia.(i)
Answer : option A
genotype of the mother XHX+
MAN ------genotype of the father XCY
nondisjunction occurred in the father
color-blind man and a woman who has a wild-type phenotype have a daughter rwith Turner
syndrome (XO) who has normal color vision.
Answer : option A
genotype of the mother X+X+
genotype of the father XCY
nondisjunction occured in the father
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