112 B 101 Lab Manu eny lketonuria (PKU) is characterized by severe intellectual
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112 B 101 Lab Manu eny lketonuria (PKU) is characterized by severe intellectual impairment due to an abnormal ne trom the presence of a breakdown product, phenylketone, in the urine and blood. Newborn babies tthe common anino acid phenylalanine within cells, including neurons. The disorder takes ery testcd at the Iwospital and, if necessary, are placed on a diet low in phenylalanine. Phenylketonuri s a recessive disorder. Mr. and Mrs. Martinez appear to be normal, but fat molecule that accumulates ormal and healthy at birth, but they they have a child with PKU. What are the genotypes of Mr. and Mrs. Martinez 5. Tay-Sachs disease is caused by the inability to break down a certain type o around nerve cells until they are destroyed. Afflicted newborns appear mo do not develop normally. At first, they may learn to sit up and stand, but later they regress intellectually impaired, blind, and paralyzed. Death usually occurs between ages three and tour and become Tay-Sachs is an autosomal recessive disorder. Is it possible for two individuals who do not have Tay-Sachs to have a child with the disorder? Explain your answer X-Linked Disorders The sex chromosomes designated X and Y carry genes just like the autosomal chromosomes. S genes, particularly on the X chromosome, have nothing to do with gender inheritance and are said to be X-linked. X-linked recessive disorders are due to recessive genes carried on the X chromosomes. Males are more likely to have an X-linked recessive disorder than females because the Y chromosome is blank for this trait. Does a color-blind male give his son a recessive-bearing X or a Y that is blank for the recessive allelle? The possible genotypes and phenotypes for an X-linked recessive disorder are as follows Females X"X8 = normal vision X*X normal vision (carrier) x X Males XY normal vision xoY color blindness color blindness An X-linked recessive disorder in a male is always inherited from his mother. Most likely, his mother is heterozygous and therefore does not show the disorder. She is designated a carrier for the disorder. Figure 11.5 shows how females can become carriers. 1. a. What is the genotype for a color-blind female? How many recessive alleles does a female inherit to be color blind? b. What is the genotype for a color-blind male?- How many recessive alleles does a male inherit to be color blind? 2. a. With reference to Figure 11.Sa, if the mother is a carrier and the father has normal vision ( .), what are the chances that a daughter will be color blind? b. A daughter will be a carrier? c. A son will be color blind? 3. a. with reference to Figure 1 1.5b, if the mother has normal vision (?? ) and the father is color blind ( ), what are the chances that a daughter will be color blind? b. A daughter will be a carrier? c. A son will be color blind? 124 Laboratory 11 Human Genetics 11-6Explanation / Answer
Answer:
4).
PKU allele= p ; Normal allele= P
Both Mr. and Mrs. Martinez are heterozygous = Pp
5).
Tay-Sachs allele = t ; Normal allele = t
Yes, when the both parents are heterozygous, they do not express the disease as the disease is autosoma recessive trait.
Tt x Tt
T
t
T
TT (normal)
Tt (normal)
t
Tt (normal)
tt (Tay-sachs disease)
X-Linked disorders:
1a).
Genotype for a color-blind female = XbXb
One recessive allele from female (mother) and another recessive allele from male (father) will be united to be colorblind in females.
1b). Genotype of colorblind male = XbY
No recessive allele is inherited from male to be inherited male. But the one recessive allele is inherited from female to be colorblind in males.
2a).
Carrier mother (XBXb) x Normla father (XBY)
XB
Y
XB
XBXB (normal daughter)
XBY (normal son)
Xb
XB Xb (carrier daughter)
Xb Y (color blind son)
If all the children are considered as 100%, the proportions are as follows
The chance of colorblind daughter = 0%
2b). The chance of carrier daughter = 25%
2c). The chance of colorblind son = 25%
3a).
Normla mother (XBXB) x colorblind father (XbY)
Xb
Y
XB
XBXb(carrier daughter)
XBY (normal son)
3b). If all the children are considered as 100%, the proportions are as follows
The chance of colorblind daughter = 0%
2b). The chance of carrier daughter = 50%
2c). The chance of colorblind son = 50%
T
t
T
TT (normal)
Tt (normal)
t
Tt (normal)
tt (Tay-sachs disease)
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