genetics 4 pool questions

b. AA Bb Cc dd

A B C d A B c d A b C d A b c d 4 different types

What is a cistron?

A section of a DNA or RNA molecule that codes for a specific polypeptide

c. What is the probability that individual II-3 is a carrier?

2/3

e. two will have galactosemia and two will not, regardless of order?

9/256 * 4 = 36/256

In humans, a dimple in the chin is a dominant characteristic. a. A man who does not have a chin dimple has children with a woman with a chin dimple whose mother lacked a dimple. What proportion of their children would be expected to have a chin dimple? Show the parents' genotypes.

(father dd) x (Dd mother) - ½ dimpled children Dd : ½ non dimpled children dd

What does KCN do?

- dissipates the potential across the bacterial plasma membrane - permits binding of the phage particles to the bacterial wall by preventing the injection of the phage DNA into the cell - Dilution of the KCN then permits infection and the lytic cycle to proceed.

d. What is the probability that individual III-1 is affected by the disease?

1 * ⅔ * ¼ = ⅙

In a mating between IV-2 and IV-5, what is the chance that the child produced would have neither syndrome?

1 - (7/16 + 1/16 + 1/16) = 7/16

b. at least one child will have galactosemia?

1 − 81/256 = 175/256

There are 4 quantities that you must measure in order to work out the recombination frequency between the two mutations. What are they?

1. concentration of unadsorbed phage from U series plates 2. total concentration from the b series of plates 3. concentration of wild type phage from K series 4. titer for all plates

an even number?

1/2 (1/6+1/6+1/6)

e. an even number on one and an odd number on the other?

1/2 x 1/2 = 1/4

In a mating between IV-2 and IV-5, what is the chance that the child produced would have alkaptonuria alone?

1/4 X 1/2 X 1/2 = 1/16

c. In a mating between IV-2 and IV-5, what is the chance that the child produced would have both nail-patella syndrome and alkaptonuria?

1/4 x 1/2 x 1/2 = 1/16

If you roll a die (singular of "dice"), what is the probability that you will roll: a. a "6"?

1/6

f. matching numbers?

1/6 x 1/6 = 1/36

b. What are the chances that their child will have cystic fibrosis?

1/9

b. What is the probability that individual II-2 is a carrier?

100%

c. Aa Bb cc Dd

2 * 2 * 1 * 2 = 8

d. Aa Bb Cc Dd

2 * 2 * 2 * 2 = 16 different types

How many genetically different eggs could be formed by women with the following genotypes? (1 point each) a. Aa bb CC DD

2 different types of gametes - A b C D + a b C D

How many different genes are we following in the phage recombination experiment? How many alleles are there of each?

2 genes, 4 alleles (rII 29, rII31) there would be two alleles of each - wildtype and mutant

A young couple went to see a genetic counselor because each had a sibling with cystic fibrosis. Cystic fibrosis is caused by a recessive disease allele, and neither member of the couple nor any of their four parents is affected. a. What is the probability that the female of this couple is a carrier?

2/3

In a mating between IV-2 and IV-5, what is the chance that the child produced would have nail-patella syndrome alone?

3/16 + 1/8 = 7/16

c. What is the probability that their child will be a carrier of the cystic fibrosis disease allele?

4/9

What is the name for this kind of test?

Complementation test

A man with a chin dimple and a woman who lacks the dimple produce a child who lacks a dimple. What is the man's genotype?

Dd

Among Native Americans, two types of earwax (cerumen) are seen: dry and sticky. A geneticist studied the inheritance of this trait by observing the types of offspring produced by different kinds of matings. He observed the following numbers: Offspring Parents Number of pairs Sticky Dry Sticky x sticky 10 32 6 Sticky x dry 8 21 9 Dry x dry 12 0 43 a. How is earwax type inherited?

Dry is the recessive phenotype and sticky is the dominant phenotype

What are the genotypes of the parents in cross [b] or [c] on page 34?

FI virgin females x dp;se males sedp, se+dp, sedp+, or se+dp+ dp;se males

Why is this ratio desirable for our purposes?

It's better to have a higher level of infection that is also controlled

People with nail-patella syndrome have poorly developed or absent kneecaps and nails. Individuals with alkaptonuria have arthritis as well as urine that darkens when exposed to air. Both nail-patella syndrome and alkaptonuria are rare phenotypes. In the following pedigree, vertical red lines indicate individuals with nail-patella syndrome, while horizontal green lines denote individuals with alkaptonuria. a. What are the most likely modes of inheritance of nail-patella syndrome and alkaptonuria?

Nail Patella syndrome will be dominant because all children that have this disorder have an affected parent Alkaptonuria will be recessive because the affected child does not have to have an affected parent

Piebald spotting is a condition found in humans in which there are patches of skin that lack pigmentation. The condition results from the inability of pigment-producing cells to migrate properly during development. Two adults with piebald spotting have one child who has this trait and a second child with normal skin pigmentation. a. Is the piebald spotting trait dominant or recessive? What information led you to this answer?

Piebald is a dominant characteristic, if the piebald was a recessive trait than they would not be able to have a normal skinned child

What are the genotypes of the parents?

Pp x Pp

Albinism is a condition in which pigmentation is lacking. In humans, the result is white hair, nonpigmented skin, and (usually) blue eyes. The trait in humans is caused by recessive alleles. Two normal parents have an albino child. What is the probability that their next child will be albino? Explain your reasoning.

Pp x Pp = pp ¼ PP Pp Pp pp

Is the disease shown in the following pedigree caused by a dominant or a recessive allele? Why? Based on this limited pedigree, do you think the disease allele is rare or common in the population? Why? (3 points)

Recessive, it has to be recessive if a child can be affected with no affected parent. This appears to be common since there are multiple unrelated individuals in this pedigree that are carriers.

By what 2 different mechanisms could mutant phages change to become capable of lysing E. coli K?

Recombination to form the wildtype phage, reversion

A mutant cucumber plant has flowers that fail to open when mature. Crosses can be done with this plant by manually opening and pollinating the flowers with pollen from another plant. When closed x open crosses were done, all the F1 progeny were open. The F2 plants were 145 open and 59 closed. A cross of closed x F1 gave 81 open and 77 closed. How is the closed trait inherited? What evidence led you to your conclusion?

The open trait is dominant, and the closed trait is homozygous recessive. When the F2 generation was crossed (Oc x Oc), gives a 3 open : 1 closed ratio or 145:59. And when the closed and open were crossed (Oc x cc) gives a 1 closed : 1 open ration or 81:77.

Which individuals must be carriers?

The parents of affected alkaptonuria children

Why are no 3:1 or 1:1 ratios present in the data shown?

The sticky parents could be either SS or Ss

If 2 different mutant rII phages are able to lyse E. coli K (our lab manual's shorthand for K12(λ)) when they co-infect it, while neither mutant can lyse it separately, what can you say about the 2 mutations?

The two mutations are complementary, the mutations occur in different genes

The 2 phage mutants we are using in lab, rII 29 & rII 31, are NOT able to lyse E. coli K when they co-infect it. What does this tell you about the 2 mutations?

They are not complementary mutations

A man with a chin dimple and a non dimpled woman produce eight children, all having chin dimples. Can you be certain of the man's genotype? Why or why not?

You can't be completely certain of this man's genotype however it is likely that he is DD since it would be more likely for all 8 of his children to inherit D from DD rather than Dd.

Why must you multiply the "titer on K" by 2 to get the recombination frequency?

To obtain the total number of recombination events, each plaque corresponds to a single recombination event

What does chloroform do?

When the lytic cycle is completed the addition of chloroform liberates any mature phage particles still inside the bacteria. It also prevents any phage DNA and immature phage particles from developing into complete phage particles.

The self-fertilization of an F1 pea plant produced from a parent plant homozygous for yellow and wrinkled seeds and a parent homozygous for green and round seeds resulted in a pod containing seven F2 peas. (Yellow and round are dominant). What is the probability that all seven peas in the pod are yellow and round?

YY rr x yy RR - Yy Rr F1 9 yellow round : 3 yellow wrinkled : 3 green round : 1 green wrinkled ( 9/16 ) ^ 7

You have just purchased a black stallion. You mate him to a red mare, and she delivers twin foals, one red and one black. Can you tell from these results how color is inherited, assuming that alternative alleles of a single gene are involved? What crosses could you do to work this out?

You wouldn't be able to tell from this cross. This is likely a test cross between a heterozygous and homozygous recessive since there was a 1:1 ratio, but you can't tell which one is which. To work this out, you can mate red horses with each other and black horses with each other. Whichever cross ends up with progeny that look like the parental phenotype would be the homozygous recessive.

Consider the pedigree that follows for cutis laxa, a connective tissue disorder in which the skin hangs in loose folds. a. Assuming complete penetrance and that the trait is rare, what is the apparent mode of inheritance?

recessive

What is multiplicity of infection (MOI)?

the ratio of phage/virus to bacteria

d. the first two will have galactosemia and the second two will not?

¼ * ¼ * ¾ * ¾ = 9/256

c. only one child will have galactosemia?

¼ * ¾ * ¾ * ¾ = 27/ 256

Galactosemia is a recessive human disease that is treatable by restricting lactose and glucose in the diet. Susan Smithers and her husband are both heterozygous for the galactosemia gene. If Susan and her husband have four children, what is the probability that: a. none of the four will have galactosemia?

¾ * ¾ * ¾ * ¾ = 81/256

a number divisible by 3?

⅙ + ⅙ = 1/3

If you roll a pair of dice, what is the probability that you will roll: (1 point each) d. two "6"s

⅙* ⅙ = 1/36