Download Which of the following genotype below shows a pure

1. Which of the following genotype below shows a pure dominant genotype?
A) Aa
B) aa
C) AA
D) AB
In this case, “pure” refers to homozygous genotype.
2. Genes that are located at the same position on homologous chromosomes are called
A) recessive B) dominant C) alleles
D) albinism
5. Which of the following would represent the sex chromosomes of a normal female?
A) XX
B) XY
C) YY
D) XXY
6. The parents of three girls are expecting another child. What are the chances the child
will be a boy?
A) 25%
B) 50%
C) 75%
D) 100%
The probability of this next child is totally unrelated to the previous children.
7. ALBINISM IN HUMANS IS INHERITED AS A SIMPLE RECESSIVE TRAIT. FOR
THE FOLLOWING FAMILIES, DETERMINE THE GENOTYPES OF THE PARENTS
AND OFFSPRING. WHEN TWO ALTERNATIVE GENOTYPES ARE POSSIBLE
LIST BOTH
A) TWO NORMAL PARENTS HAVE FIVE CHILDREN, FOUR NORMAL, AND
ONE ALBINO
Since the parents had one albino parent, we know that each parent must have been a
carrier. Therefore, each parent was Aa. If you do the cross between Aa and Aa, you
will find that you will end up with 1 AA, 2 Aa, and 1 aa. The AA and Aa children are
normal while the aa child is an albino.
B) A NORMAL MALE AND ALBINO FEMALE HAVE SIX CHILDREN, ALL
NORMAL
Since no albino children were produced, we will assume that the male is AA and the
albino female is aa. Therefore, all children were Aa.
C) A NORMAL MALE AND AN ALBINO FEMALE HAVE SIX CHILDREN, THREE
NOTMAL AND THREE ALBINO
Here, the normal male must be a carrier. Therefore, he’s Aa while the albino female is
aa. If you cross these in a typical Punnett Square, you will end up with a 1:1 ratio of
Aa (normal children) and aa (albino children).
D) CONSTRUCT A PEDIGREE OF THE FAMILIES IN (B) AND (C) ASSUME THAT
ONE OF THE NORMAL CHILDREN IN (B) MARRIES ONE OF THE ALBINO
CHILDREN IN (C) AND THAT THEY HAVE EIGHT CHILDREN.
You can draw the pedigree itself, but here’s the basics you’ll need. The normal child
in (B) is Aa, right? He marries an albino (aa). If you do the cross, you will predict a
1:1 ratio of normal children to albino children. If you have eight children, you can
predict 4 of each type.
10. Using the forked-line, or branch diagram, method, determine the genotypic and
phenotypic ratios of the trihybrid crosses.
I’ll do one for you and then you can do the others. After all, we are not supposed to just
do your homework, but tutor you in how to do it, right?
(a) AaBbCc x AaBBCC
How may gametes will AaBbCc produce? Every possible combination must be taken into
consideration. Here they are: ABC, ABc, AbC, Abc, aBC, aBc, abC, abc. There are 8.
There are no other combination.
Now, we do the same thing for the AaBBCC parent. There are only 2 possible gametes
here: ABC and aBC.
Now, we do the cross:
ABC
ABc
AbC
Abc
aBC
aBc
abC
abc
ABC AABBCC AABBCc AABbCC AABbCc AaBBCC AaBbCc AaBbCC AaBbCc
aBC AaBBCC AaBBCc AaBbCC AaBbCc aaBBCC aaBBCc aaBbCC aaBbCc
Now, group them genotypically:
AABBCC
AABBCc
AABbCC
AABbCc
2 AaBBCC
2 AaBbCc
2 AaBbCC
AaBbCc
AaBBCc
aaBBCC
aaBBCc
aaBbCC
aaBbCc
Now, group them phenotypically:
All of these will have the dominant genotype for the three genes. They will look
identical:
AABBCC
AABBCc
AABbCC
AABbCc
2 AaBBCC
2 AaBbCc
2 AaBbCC
AaBbCc
AaBBCc
These will be different. They will have the dominant phenotype for B and C, but the
recessive phenotype (“a”) for the A gene.
aaBBCC
aaBBCc
aaBbCC
aaBbCc
Therefore, the phenotypic ratio is 12:4 or reduced to 3:1.
(b) AaBBCc x aaBBCc
(c) AaBbCc x AaBbCc
11. All of the sperm from one human male are generically identical
A) True
B) False (meiosis produces variation, not clones)
12. The A and B antigens in humans may be found in water-soluble form in secretions,
including saliva, of some individuals but not by others. The population thus contains
“secretors” and nonsecretors”. The following inheritance patterns have established that
this trait is inherited:
all secretors
all nonsecretors
all secretors
F2= ¾ secretors:
¼ nonsecretors
How is the trait inherited? Determine the genotypes of all parents and offspring. (address
the above pattern)
The real useful data is when secretors and nonsecretors were crossed. All children ended
up being secretors. This seems to suggest that secretors is dominant. Therefore, we use
the S for secretors and the s for nonsecretors. Therefore, secretors are either SS or Ss,
while nonsecretors are ss. In the fourth cross, when we cross the F1 secretors resulting
from the third cross (Ss), we do get the 3:1 phenotypic ratio as predicted. Do the cross
and you should see that. The ¾ secretors are 1 SS and 2 SS while the ¼ nonsecretors are
ss.
Secretor x secretor
Nonsecretors x nonsecretors
Secretor x nonsecretor
F1 secretors
13. Why is poly dT an effective primer for reverse trancriptase?
It binds really well to the poly A tail in mature mRNAs.
14. cDNA can be cloned into vectors to create a cDNA library. In analyzing cDNA
clones. It is often difficult to find clones that are full length, that is, extend to the 5’ end
of the mRNA. Why is this so?
Reverse transcription starts from the 3’ end (where the poly A tail is). As a result, many
of the enzymes fall off the template before they reach the 5’ end.
15. Acridine dyes induce frames shift mutations. Is such a mutation likely to be more
detrimental than a point mutation where a single pyrimidine or purine has been
substituted?
Frame shifts are very bad because they throw off the entire reading frame downstream
from the mutation. Therefore, every amino acid translated from that point on will be
wrong. On the other hand, point mutations may only result in a single amino acid
substitution. These are not nearly as disasterous.
16. Linked genes can separated on a chromosome by the process of
A) crossing over
B) mutation C) linkage
D) no disjunction
19. What molecule brings amino acids to the ribosome in the cytoplasm during protein
synthesis? tRNAs
27. Looking at the codon sequence here, how would this mutation be classified?
ATC-TGC-GGA
DNA sequence before mutation
ATT-CTG-CGG
DNA sequence after mutation
A) insertion B) deletion
C) substitution
D) none of these.
28. Why is the above answer correct?
I’ve highlighted the inserted base. Look at it carefully. If you remove that “T” and put the
second sequence back together again, you will have the first sequence.
29. How many participants are involved in a TRIO paternity test?
A) 5 B) 4 C) 3 D) NONE of the above
You’re testing the mother, the child and the suspected father.
30. When taking a buccal swab collection, were are samples taken from?
A) Blood
B) Finger
C) mouth
D) urine
31. When taking samples from a swab DNA is been gathered when plenty of saliva is
present. (This sentence does not make grammatical sense; therefore, I cannot answer it.
Please rewrite it or check the original. Something is wrong here.)
A) true
B) false
32. Typically in a buccal swab test, what is the standard amount of time?
Time for what? This question is not specific enough and, once again, cannot be answered.
Please get some clarification from teacher as to what time is being questioned.
A) 10 second B) 30 seconds C) 5 seconds D) any amount of time
33. What organization does the acronym AABB stand for? “Arizona Association of
Business Brokers”? No, actually in this context it is “American Association of Blood
Banks”
34. Down syndrome results from genetic mutation called
A) histamine B) nondisjunction
C) substitution
D) insertion
35. How many nitrogen bases are needed to code for one amino acid?
A) 1 B) 2 C) 3 D)4
36. The biological process of mRNA copying the original DNA molecules is called
A) translation B) transcription
C) hydrolysis D) dehydration synthesis
MATCHING
Match the genetic cross of the parents on the left with the genotypes on the right of the
offspring most likely to be produced from the cross. You may use an answer choice more
than once or not at all.
(Write answers on the lines provided below)
GENETIC CROSS
PREDICTED OFFSPRING
37. BB x bb _C_
A. 25% BB, 50% Bb, 25% bb
38. Bb x Bb _A_
B. 100% bb
39. BB x BB _E_
C. 100% Bb
40. bb x bb _B_
D. 50% Bb, 50% bb
41. Bb x bb _D_
E. 100 BB
F. 75% Bb, 25% bb
Do the crosses to confirm.
Match the genetic cross of the parents on the left with the phenotypes on the right of the
offspring most likely to be produce from that cross. You may use an answer choice more
than once of not at all.
GENETIC CROSS
PREDICTED OFFSPRING
42. TT x tt
_C__
A. 100% short
43. tt x Tt
_E__
B. 75% tall, 25% short
44. Tt x Tt
_B__
C. 100% tall
45. TT x TT _C__
D. 25% tall, 75% short
46. tt x tt
_A__
E. 50% tall, 50% short
48. Which of the following conditions would not be desirable in genetic study?
A) Testing organisms that produce large number of offspring
B) Testing organism that are small in body size.
C) Testing organisms that have a long life cycle.(because it would take too long to do
multigenerational studies.)
D) Testing organism that have large chromosomes.
49. The “address” of a gene on a chromosome is called its _______locus_______.
50. Alternative forms of a gene (e.g. short and tall in garden peas, t and T) are called
____________alleles____________.