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Unformatted text preview: Biological Sciences 101 -002
Dr. Mark Sanders
Winter Quarter 2010 First Midterm Exam Answer Key — FORM C
1A. 28 chromatids
1B. 7 chromosomes
,2 M4 1C. 28 chromatids
1D. 10 picograms
1B. 5 picogram ( 1 2A. The F1 are wild-type because the plants are heterozygous for the ﬂower color
[ mutation. The 3:1 F2 ratio is consistent with inherited variation of a single gene causing
Z ﬂower color mutation.
3 4 2B. Genetic complementation
(6 z 2C. Both mutant lines carry mutations of the same gene.
< 2D. The F1 progeny are dihybrid and are expected to produce F2 progeny in the ratio 9/16
yellow to 7/16 white. The F2 progeny will be r+_d+_ (yellow) 9/16, r+_dd (white) 3/ 16,
L z rrd+_ (white) 3/16, rrdd (white) 1/16. I 6,4, 3A. Cross 1: T tAa x T tAa Both parent tall, axial
Cross 11: TTAa x T-Aa Both parents tall, axial
3B. )8 = (92-90)2/9o + (28—30)2/30 + (29—30)2/30 + (11-10)2/10 = 0.310
//M /- Df = 3; P value > 0.90, independent assortment between the gene is not rejected
on the basis of the P value bbing greater than 0.05. \ l \/ / 4A. Tall parent Short parent Tall F1 Lane 1 2 3 Kb z 4B. All three band patterns are possible. Homozygous tall will have a single band of 3.4
:5 kb, homozygous short will have a single band of 2.2 kb and heterozygous (tall) will have z both bands. z 5A. The mutant allele is dominant. The heterozygous organism nor the mutant
Z homozygote fall short of producing sufﬁcient enzymatic activity to produce the wild-type
1 z phenotype and both are phenotypic mutants. SB.
Unaffected Unaffected Affected Affected
Female Male Heteroz ous Homo ous Heteroz ous Homoz ous 6A. Each prospective parent has a 2/3 chance of being a carrier (conditional probability).
wait The chance the ﬁrst child has galactosemia is (2/3)(2/3)(l/4) = 4/36 = 1/9 63. If the ﬁrst child has galactosemia, then the parents are carriers. The second child is
the product of a heterozygous mating and has a 1A chance of having the disease. “,3” 6C. If the parents are carriers, there is a ‘A chance of disease in each subsequent child and
<9- a 3A chance of being disease-free. The probability that child 2 and child 3 are disease-free
is (3/4)(3/4) = 9/16.
7A. The trait is most likely X-linked recessive. The preponderance of males, the
l “l uniformity of affected males from an affected mother in generation 11, and the ease of X-
‘pay/lb linked recessive explanation of affected males in generations III and IV all argue in favor
l of this mode of inheritance. Q 4/ L 7B. Woman A is most likely GG. Her mate is gY. » 7C. The man C is GY. The woman B has a 1/: chance of being a carrier (Gg). Her
; maternal grandmother was a carrier and there is a ‘/2 chance of passing the X-linked
aub recessive to Cs mother and another 1/2 chance her mother passes the allele to C. The son
1 #3 cm and c has a (1/2)(1/4) = 1/8 chance of being affected. 8. The format of this answer is illustrated in Figure 3-8, page 104 of the text. ...
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- Fall '09
- Genetics, ﬁrst child, galactosemia, Midterm Exam Answer, parent Tall F1