Week 2 Optional Quiz Modifications of Mendel Attempt review PDF

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Quiz practice based on week 2 - useful for mcq exam (full question and answers)...

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Started on State Completed on Time taken Marks Grade Question 1 Not answered Marked out of 1.00

Sunday, 14 March 2021, 8:19 PM Finished Sunday, 14 March 2021, 8:19 PM 14 secs 0.00/11.00 0.00 out of 100.00

Two people with recessive deafness marry and have ten children, all of who can hear. The most likely explanation of this is: Select one: a. Incomplete Penetrance b. Pleiotropy c.

Codominance

d. Mutation e. Complementation f.

Sex Limitation

g. Incomplete dominance

Your answer is incorrect. There are many dierent loci aecting hearing – some aect outer ear, some the malleus, incus and stapes of the middle ear, some the cilia in the inner ear, some nerve connections to the brain or the brain itself. A break in the chain for any of them will lead to deafness. We can model it thus: In order to hear, one must have at least one capital allele: Hearing phenotype: Sound > A > B > C > D > E > F > Perception of sound Genotypes: First individual: AA > BB>cc X DD X EE X FF – deaf (homozygous mutation at the C locus) Second individual: AA > BB > CC > DD > eeX FF – deaf (homozygous mutation at the E locus) Ospring: AA > BB > Cc > DD > Ee > FF – can hear (heterozygous at the C and E locus) ie complementation between C and E locus gives normal hearing as Cc and Ee heterozygotes have functioning steps in auditory chain. As a result, two deaf people often have hearing children as each parent is homozygous for a recessive abnormality at a dierent locus and child is double heterozygote. The correct answer is: Complementation

Question 2 Not answered Marked out of 1.00

In a sheep breeds the presence of a horn is determined at a single loci which shows complete dominance. In Dorset Horned Sheep male and female homozygous dominant individuals (h +h +) both have horns. In Suolk sheep male and female homozygous recessive individuals (hh) are both hornless. A cross between the two breeds found that all males were horned and all females were hornless. What is the best explanation for this? Select one: a. Complementation

b. Mutation

c.

Incomplete dominance

d. Sex limitation e.

f.

Codominance

Pleiotropy

g. Incomplete Penetrance

Your answer is incorrect. In some organisms, traits are inuenced by the sex of the organism. In other words, males and females which are genotypically the same at a particular trait loci, give dierent expressions of the same trait. In this cross a h+h+ individual was crossed with a hh individual. The ospring were all h+h (ie one copy of the dominant allele and one copy of the recessive allele). The phenotypic expression of the h+h genotype deferred in males and females; males had horns and females were hornless. The same allele had dierent expressions in male and female individuals. This is an example of sex limitation. The correct answer is: Sex limitation

Question 3 Not answered

A cross was made between snapdragons with red and white owers, and the resulting ospring were all pink. When the pink F1 snapdragons were crossed, the following ospring were produced: 125 red, 242 pink, and 123 white.

Marked out of 1.00

Assume that one of the pink F2 plants is crossed to a white plant. Of the 400 ospring produced, what would be the expected number of pink plants? Answer: 

Three distinct phenotypes are produced from the cross between red and white owers, with the new phenotype, pink, being an intermediate of the two parental phenotypes. This suggests that the pattern of inheritance is incomplete dominance. With incomplete dominance, the heterozygote is generally an intermediate of the two homozygotes. We are not told which loci is dominant and recessive, but this information is not needed. This can be claried by looking at the crosses: Parental Phenotypes: Red X White Parental Genotypes: R1R1 X R2R2 All F1 ospring are Pink: R1R2 The cross between the F1 snapdragons will therefore produce the following genotypes:

This is the classic 1:2:1 ratio associated with incomplete dominance. The pink individuals from the F2 (R1R2) are then crossed with a white plant. The cross looks like this:

The ratio of Red:Pink:White = 0:1:1 Of the 400 plants produced we would therefore expect to observe half of the ospring to be pink, ie 200. The correct answer is: 200

Question 4 Not answered Marked out of 1.00

Found on chromosome 4 in humans, two blood group alleles L M and L N are equally expressed in heterozygous individuals. Such heterozygous individuals are said to be in the MN blood group. Assume that a man and a woman, both in the MN blood group, have two children. What is the probability that both children will be in the MM blood group? Give your answer in the form of a fraction, for example 1/2. Answer: 

Since the rst sentence refers to two alleles, this must be a monohybrid cross situation. Given that in heterozygotes, the L M and L N alleles are equally expressed, there is good evidence that this is an example Codominance. The crosses are therefore:

P(One child will be in the MM blood group) = 1/4 P (both children will be in the MM blood group) = 1/4*1/4 = 1/16 The correct answer is: 1/16

Question 5 Not answered Marked out of 1.00

Plumage in ducks is determined to some extent by a multiple allelic series of alleles named restricted (M R= white on wing fronts), mallard (M = wild-type), and dusky (m d = darker plumage). M R is dominant to M, which is dominant to m d. A series of matings between two mallard ducks produced the following ospring: 34 mallard and 11 dusky ducks. What are the likely genotypes of the two mallard parents? Select one: a. MR md X M md

b. M md X M md c.

M md X M M

d.

md md X M M

e. M M X M M

Your answer is incorrect. This is an example of multiple alleles; we have 3 dierent alleles which can occupy the same loci. Given that MR is dominant to M we know that this allele is not in play here, so in eect this is just a monohybrid cross. If 34 mallard and 11 dusky ducks are produced, this approximates a 3:1 ratio or a typical monohybrid cross between two heterozygotes. Given that to express the dusky phenotype requires two copies of the md allele, the cross to produce the results would be: M md X M md

The correct answer is: M md X M md

Question 6 Not answered Marked out of 1.00

Coat colour in mice is determined to some extent by two independently assorting but interacting gene loci. The dominant allele (B) at one locus allows a colourless precursor to form black pigment while its recessive allele (b) inhibits pigment formation entirely and gives the albino phenotype when homozygous. At the second locus the dominant A allele allows the pigment to be distributed in the agouti pattern. In the absence of the A allele (ie when the genotype is aa), only the solid black pigment pattern occurs in the fur. Assume crosses are made between two fully heterozygous mice. What phenotypes would be expected and in what numbers if 160 ospring are produced? Select one: a. Agouti: 90 Black: 30 Albino:40

b.

Agouti: 120 Black: 20 Albino: 20

c.

Agouti: 120 Black: 30 Albino: 10

d. Agouti: 53.3 Black: 53.3 Albino: 53.3

e. Agouti: 40 Black: 40 Albino: 80

Your answer is incorrect. A cross between two fully heterozygous mice would be symbolised as shown below. This would be a typical dihybrid cross that would normally yield a 9:3:3:1 ratio if no gene interaction were to occur. However, because of gene interaction, a modied phenotypic ratio will result. Cross: Aa Bb X Aa Bb

Red: Agouti Blue: Albino Black: Black As long as the B allele is present, the black pigment will be produced. If an A allele is present, the black pigment will be distributed in the agouti patterning. If no B allele is present, the mouse will be albino, regardless of the presence of the A allele. If the B allele is present, but no A allele is present, the mouse only produces the black pigment with no additional patterning. The ratio of Agouti: Albino: Black is therefore 9:4:3. As such if there are 160 ospring, we would expect to obtain: Agouti: 9/16 * 160 = 90 Albino: 4/16 * 160 = 40

Black: 3/16 * 160 = 30 The correct answer is: Agouti: 90 Black: 30 Albino:40

Question 7 Not answered Marked out of 1.00

Coat colour in the mouse serves as a good example of epistasis. The normal wild-type colour is agouti, which is a grizzled (greyish) appearance of the fur due to alternating light and dark bands of individual hairs. This phenotype requires at least one dominant allele at each of the dierent loci (A_B_). Whenever the b allele is homozygous recessive, the albino trait will occur. Recessiveness at the a locus with at least one dominant B allele at the other locus yields a black coat. What is the expected ratio (Agouti: Black: Albino) for the progeny of a backcross between Aa Bb and aa bb? Select one: a. 9:4:3 b. 2:2:3 c.

1:2:3

d. 1:1:2 e. 1:1:1

Your answer is incorrect. The solution can be found using a punnet cross:

The Expected ratio Agouti: Black: Albino is 1:1:2 The correct answer is: 1:1:2

Question 8 Not answered Marked out of 1.00

In horses, assume that WW is chestnut, Ww is palomino, and ww is white. What ospring would you expect from a white stallion and a palomino mare? Select one: a. 1:1:1 ratio Palomino: Chestnut: White b. 1:2 ratio Palomino: Chestnut c.

1:1 ratio Palomino: Chestnut

d. 1:2 ratio Palomino: White e. 1:1 ratio Palomino: White

Your answer is incorrect. Horse coat genetics is a classic example of incomplete dominance. Here, the heterozygote is an intermediate value of the two heterozygotes.

We would therefore expect a1:1 ratio of Palomino (Ww) to white (ww). The correct answer is: 1:1 ratio Palomino: White

Question 9 Not answered Marked out of 1.00

This question involves the three types of loci found in the agouti mice. When two mice with the genotype of AaBBCC and AABbcc cross together, what is the possible colour of the F1 ospring? Notes: A-B- gray A-bb yellow aa B-black aa bb cream C- normal cc albino Select one: a. Only black is possible, as the zygotes containing the aaBBCc genotypic combination are not viable

b. All of the ospring will be grey (1:1:1:1 ratio for all the possible gene combinations) c.

d.

Half of the litter will be black, while the other half will be grey

Half of the litter will be grey, and the other half will be cream

e. Half of the litter will be cream, while the other half will be black

Your answer is incorrect. This question involves the concept of gene interaction and in the lectures, we were introduced to at least ve of the example loci which regulate the actual appearance of the agouti mice together. Again: A-B- gray A-bb yellow aa B-black aa bb cream C- normal cc albino As each of the locus is independent, we tackle this question by drawing crosses individually. e.g. Aa x AA will give a possible combination of half AA and half Aa Bb x BB: half BB and half Bb. CC x cc: All Cc (i.e. no Albino in this cross, since albino is recessive) Therefore, we could see that there will be 4 possible combinations: A BBC

AABbCc AaBbCc AABBCc And: All the ospring have the A-B- genotypic combination, they will all be grey. ** There are not any known genotypic combinations of the loci A, B, C alleles that will lead to fatality before birth in agouti mice. The correct answer is: All of the ospring will be grey (1:1:1:1 ratio for all the possible gene combinations)

Question 10 Not answered

Manx cats with two recessive alleles (tt) have a tail. Heterozygous Manx (Tt) Cats lack a tail. Homozygous Dominant (TT) cats die before birth.What proportion of Manx cats do not have a tail in the living population, when the following two crosses occur:

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i) Manx cat with a tail crossed with a Manx cat without a tail ii) Manx cat without a tail crossed with a Manx cat without a tail Select one: a. i) 1/2 ii) 2/3

b. i) 1/2 ii) 1/2

c.

i) 1/2 ii) 3/4

d. i) 2/3 ii) 1/4 e.

i) 2/3 ii) 1/3

Your answer is incorrect. In the rst cross i): 1/2 probability ospring will be a heterozygous cat (Tt) without a tail 1/2 probability ospring will be a homozygous recessive cat with a tail (tt) In the second cross ii) 1/4 probability ospring will be homozygous dominant cat (TT) and will therefore die before birth. Homozygous dominant cats are NOT members of the living population. Therefore, from the ospring that will survive and are part of the living population: 1/3 probability ospring will be a homozygous recessive cat with a tail (tt) 2/3 probability ospring will be a heterozygous cat without a tail (Tt) The correct answer is: i) 1/2 ii) 2/3

Question 11 Not answered Marked out of 1.00

Labrador retrievers come in 3 colours: black, brown and cream/yellow. The allele B codes for dark pigment including nose and skin (e.g. inside the mouth, around eyes), BB or Bb is a black lab with black pigment, whereas bb is brown (b is recessive) and results in brown pigment on the skin. E allele on another loci codes for type of melanin calledeumelanin.An EE lab will produce a dark pigmented labrador (bb E_-brown, B_ E_ -black), whereas ee would result in a yellow lab with a yellow melanin production (NOT to be confused with albino, still has melanin). However, bb ee labs are yellow labs with BROWN pigment on nose and skin and are DISQUALIFIED from dog shows including Crufts. A lab breeder has crossed his 2 beloved dogs with the following genotypes: bb Ee x Bb Ee What is the predicted number of possible puppies in the ock that may be disqualied from dog shows, if 16 puppies were born?

Answer: 

This can be worked out from the cross:

1/8 of the puppies will have yellow fur, brown skin and would be disqualied. 1/8 X 16 = 2 The correct answer is: 2

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P REV IO US ACTIV ITY Lecture 4: Modications of Mendel II

NE XT ACTIV ITY Genetics for the Perplexed: Modications of Mendel I

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