1) In pea plants, round peas (R) are dominant to wrinkled peas (r).You do a test
ID: 300637 • Letter: 1
Question
1) In pea plants, round peas (R) are dominant to wrinkled peas (r).You do a test cross between a pea plant with wrinkled peas (genotype rr) and a plant of unknown genotype that has round peas. You end up with three plants, all which have round peas. From this data, can you tell if the round pea parent plant is homozygous dominant or heterozygous? If the round pea parent plant is heterozygous, what is the probability that a random sample of 3 progeny peas will all be round?
a) The data set is too small to predict the genotype of the round pea plant. Assuming that the unknown parent is heterozygous, the probability of having only round pea plants from a random sample of 3 progeny will be 1/8 . b) The genotype of the unknown round pea plant is Rr. Assuming that the unknown parent is heterozygous, then the probability of having only round pea plants from a random sample of 3 progeny will be 1/4 . c) The genotype of the unknown round pea plant is Rr. Assuming that the unknown parent is heterozygous, the probability of having only round pea plants from a random sample of 3 progeny will be 1/2 . d)The data set is too small to predict the genotype of the round pea plant. The probability of having only round pea plants from a random sample of 3 progeny will be 1/6 .
2) The most common form of hemophilia affects 1 out of every 5,000 male births worldwide, but the condition is much rarer in females. Explain why this is the case.
a) Females need two mutated X chromosomes to be hemophilic.
b) Females need one mutated X chromosome to be hemophilic.
c) Females do not inherit mutated X chromosomes.
d) Females need two mutated X chromosomes to not be hemophilic.
3) A true-breeding white mouse (bb) is repeatedly mated with a true-breeding brown mouse (BB). If the dominant phenotype is brown, what genotype will the F1 generation be? What percentage of the F2 generation will be brown if individuals from the F1 generation are mated?
a) Bb, homozygous; 100%
b) Bb, heterozygous; 75%
c) BB, homozygous; 50%
d) Bb; heterozygous 50%
In a study on cell division, researchers culture synchronously dividing human cells with thymidine. This causes the cells to arrest at the G1/S boundary. The cells are then placed in medium lacking thymidine, which releases the block, and the cells begin to divide again. Starting with Sample A and ending with Sample D, the DNA content of the cells is measured at different times after thymidine is removed. Results for four samples (A-D) are shown in the graph. What would be the expected results for the cells that were kept in medium lacking thymidine past the sample D time point (additional time points E and F)?
2 1 0 Sample Sample Sample SampleExplanation / Answer
Q1) Here" R " that is round pea is dominant over " r " wrinkled pea. and a test cross is done between :"rr" wrinkled pea and a plant of unknown genotype which has round peas , that gave rise to 3 round pea plants.
Answer- will be C , because here the probability of getting round pea plants is 1/2 or 50 % .
Explanation- In a test cross the cross is done between a heterozygous plant and a homozygous either recessive or dominat character .
here the cross is done between Homozygous recessive that is "rr" and Heterozygous round pea plant so the genotype will be "Rr"
In the punnett square it is seen when" Rr" crossed with "rr"
The geotype of the progeny is "Rr" and "rr " and the chance of getting a round pea is 50% or 1/2
here 1/2 of the genotype are heterozygous round and half are homozygous wrinkled.
Q2) The most common form of hemophilia affects 1 out of every 5,000 male births worldwide, but the condition is much rarer in females. Explain why this is the case.
Answer- A- Females need two mutated X chromosomes to be hemophilic.
Explanation-
Haemophilia is a sex linked recessive disorder. The mutated gene is found on X chromosome , which comes either from father or mother in a female , and in a male it comes from mother.
a person to be Haemophilic , he needs to have two mutated gene of Haemophilia , as it is recessive sex linked disease , so he got to inherit one faulty X gene from father and one faulty gene from Mother .As in case of female if a normal X gene present in one locus than it becomes dominant on the haemophilic gene , and the female is a carrier but do not show haemophilia. But in case of male as it requires only 1 faulty X gene to have the disease , so disease is rarer in female than male. usually female are the carriers of the disease.
Q3) A true-breeding white mouse (bb) is repeatedly mated with a true-breeding brown mouse (BB). If the dominant phenotype is brown, what genotype will the F1 generation be? What percentage of the F2 generation will be brown if individuals from the F1 .
Answer:- B- Bb, heterozygous; 75%
Explanation:- Here bb is crossed with BB
in punnett square when "bb" crossed with "BB" it would be
From the above punnett square it is found that the genotype of F1 generation will be Bb.
when the offsprings of F1 generation were crossed that is " Bb " crossed with "Bb" the F2 generation will be
In F2 generation 75% of the offspring will be Brown having genotype "BB" and "Bb" as brown that is "B" is dominat over b.
Q4)
In a study on cell division, researchers culture synchronously dividing human cells with thymidine. This causes the cells to arrest at the G1/S boundary. The cells are then placed in medium lacking thymidine, which releases the block, and the cells begin to divide again. Starting with Sample A and ending with Sample D, the DNA content of the cells is measured at different times after thymidine is removed. Results for four samples (A-D) are shown in the graph. What would be the expected results for the cells that were kept in medium lacking thymidine past the sample D time point (additional time points E and F)?
Answer- C-
Explanation- Answer will be C because thymidine is removed, after the D time point , as in the sample from A to D it is seen , DNA content was low in sample A but doubled from sample B onwards, it takes some time to enhance the production of DNA content , after the removal of thymidine. So wjen DNA removed, post sample D , the DNA content was low in sample E and but started doubling in F.
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