by cstmweb

Question posed by Sarah Boo: GENETICS QUESTION!!!?
A mother with cystic fibrosis marries a man who is a carrier of cystic fibrosis. What are the chances their child will be a carrier of the disease?

Most detailed answer:

Answer by DAVID HENDER
50% chance the child will be a carrier and 50% chance they will have cystic fibrosis as it is a recessive disorder.
So 50%

Do you know better? Why not leave your own answer in the comments below!

Technorati Tags: GENETICS, question

by k-ideas

A question from s3jealousy: Genetics Problems w/ PUNNET SQUARES?
Please I am so lost with this.

1. In humans, blue eyes are recessive to brown eyes. Draw a punnet square showing the expected ratios of chilren born to a blue-eyed woman and a brown-eyed man whose mother was blue eyed.

2. In cats, short hair is dominant to long hair. A shorthaired tomcat is mated to a longhaired female. She has eight kittens: 6 shorthaired and 2 longhaired. Diagram the cross. How do the results compare to the expected ratios?

3. Cystic fibrosis is recessively inherited trait. Two normal people have a child who has cystic fibrosis. What was the probability that this couple produces a child with cystic fibrosis. What is the probablility that this couple will have two children in a row afflicted with cystic fibrosis?

4. The gene for red/green eye color vision is sex-linked. Red/green color blindness is recessive. Would the children of a color-blind woman and a man of normal color vision be color-blind?

5. The chestnut color of horses is due to a recessive gene, while the dominant allele results in a black coat. In terms of gaits, a horse may be a trotter (dominant) or a pacer (recessive). By means of a Punnet Square, show the offspring which could result from a cross between a black trotter whose mother was a chestnut pacer and a black pacer whose father was chestnut.

6. Listed below are the progeny from two crosses between a species of flowering plants. What pattern of inheritance does this cross illustrate? Determine the genotypes of the parents and their progeny. Draw a punnet square to validate your answers. Show all work.
– Cross 1: blue-flowered plant X white-flowered plant gives F1: all pale-blue flowered
– Cross 2: pale-blue F1 X pale-blue F1 gives F2: 27 blue, 49 pale-blue, and 24-white flowered plants.

Most detailed answer:

Answer by Dina Felice
1. bb (woman) x Bb (man)

___ B | b
b | Bb | bb

2. Ss x ss

___ S | s
s | Ss | ss

We would expect 50-50, so 75-25 is unexpected. It is as if the female was also heterozygous, instead of homozygous long.

3. 25% or 1/4.
6.25% or 1/16

4. The sons would all be colorblind, none of the daughters would be, but all the daughters would be carriers.

5. BbTt x Bbtt

You would see 4 possible phenotypes, black trotters, black pacers, chestnut trotters and chestnut pacers, but it is 32 squares, so I’m not going to write it out.

6. This is incomplete dominant. You are crossing 2 true breeding flowers, then getting all heterozygous offspring, then normal 1:2:1 ratio in F2.

Provide your own answer to this question below!

Technorati Tags: GENETICS, problems, PUNNET, SQUARES

A question from Helen: Genetics Problem ?????????????
Cystic fibrosis is a serious genetic disease caused by an autosomal recessive allele (C= normal and c= cystic fibrosis). This trait is not rare amongst humans; approximately 1 in 25 people in human populations are carriers (i.e. heterozygous for this allele). A woman and her husband (neither of whom have the disease) are considering having children, but the woman’s brother (there were only two children in the family) has cystic fibrosis. On the other hand, the woman’s parents and her maternal grandparents (i.e. her mother’s parents) did not have the disease.

a. Draw a partial pedigree showing all of the individuals mentioned and give their genotypes as fully as possible (hint: there may be uncertainty for some).

b. What is the33 probability that the first child produced by this couple will have cystic fibrosis?

No 1 answer:

Answer by Elisa
use a punnett square. its not that difficult.

Do you know better? Why not leave your own answer in the comments below!

Technorati Tags: GENETICS, problem

by k-ideas

A question asked by edgygirl: Genetics question for Biology?
imagine that you are a genetic counselor, and a couple planning to start a family comes to you for information. Charles was married once before, and he and his first wife had a child with cystic fibrosis. The brother of his current wife, Elaine, died of cystic fibrosis. What is the probability that Charles and Elaine will have a child with cystic fibrosis. Neither Charles or Elaine has cystic fibrosis.
I know the answer is 1/6 but how do you get that?
how many punnet squares do i draw? and put what on each of the two sides. thats what i need help with.

Most detailed answer:

Answer by Yote
do a punnet square!

Do you know better? Why not leave your own answer in the comments below!

Technorati Tags: Biology, GENETICS, question

by cstmweb

Question by edgygirl: Genetics question for Biology?
imagine that you are a genetic counselor, and a couple planning to start a family comes to you for information. Charles was married once before, and he and his first wife had a child with cystic fibrosis. The brother of his current wife, Elaine, died of cystic fibrosis. What is the probability that Charles and Elaine will have a child with cystic fibrosis. Neither Charles or Elaine has cystic fibrosis.
I know the answer is 1/6 but how do you get that?
how many punnet squares do i draw? and put what on each of the two sides. thats what i need help with.

No 1 answer:

Answer by Yote
do a punnet square!

What do you think? Leave you answer below!

Technorati Tags: Biology, GENETICS, question

by k-ideas

A question asked by damigurl05: Genetics Pt. 1?
A man and a woman are both heterozygous for Cystic Fibrosis, an autosomal recessive disease. They have 16 pre-embryos screened for the disease. How many will be homozygous dominat, homozygous recessive and heterozygous. How many will be affected, how many will be carriers, and how many will not be affected?

Best answer:

Answer by TinklePuff741
C – Normal allele (dominant)
c – Cystic fibrosis allele (recessive)

Affected genotype: cc (homozygous recessive)
Carrier genotype: Cc (heterozygous / carriers)
Unaffected genotype: CC (homozygous dominant)

Man: Cc
Woman: Cc

Probability of embryos having:
CC genotype = 0.25
Cc genotype = 0.50
cc genotype = 0.25

Multiply the probability by the whole number (16) of embryos to get the predicted number of embryos with each genotype:
CC = 0.25 x 16 = 4
Cc = 0.50 x 16 = 8
cc = 0.25 x 16 = 4

Agree or disagree? Leave your own thoughts below.

Technorati Tags: GENETICS

by cstmweb

Question posed by Ashley P: genetics help please?
if one in every 22 people in the U.S is a carrier for Cystic fibrosis(autosomal recessive disease) what proportion of the U.S population would be expect to have c.f.?

The No 1 answer:

Answer by gotaprofquestion
1 in 22 x 1 in 22 = 1 in 484

If you know better then please let us know below.

Technorati Tags: GENETICS, help, PLease

by mcbill

Question posed by Ashley P: genetics help please?
if one in every 22 people in the U.S is a carrier for Cystic fibrosis(autosomal recessive disease) what proportion of the U.S population would be expect to have c.f.?

My chosen answer:

Answer by gotaprofquestion
1 in 22 x 1 in 22 = 1 in 484

What do you think? Leave you answer below!

Technorati Tags: GENETICS, help, PLease

by cstmweb

Question posed by Sarah H: genetics mathy question?
please help i really need to know!

if 1 in 50 alleles are defective for cystic fibrosis. what proportion of the live births will have the disease?

another question:
if 1 in 20 are heterozygous for a recessive condition then what are the chances of an individual in that population being born with the condition.

really stuck and that is as much i know about the question :s

The top answer:

Answer by Dink87522
1. 2% will have the disease.
2. 5% chance.

I don’t quite understand you’re question so those may be wrong.

Do you know better? Why not leave your own answer in the comments below!

Technorati Tags: GENETICS, mathy, question

by cstmweb

A question from Tigerlily2682: Part 1: Genetics – From Genes to Proteins, Mutations (Chapter 10)?
Overview: Genetic information in DNA is transcribed to RNA and then translated into the amino acid sequence of a Protein.

A) Step 1 – Transcription: During the process of transcription, the information in the DNA codons of a gene is transcribed into RNA.

Suppose that gene X has the DNA base sequence 3’-TACCCTTTAGTAGCCACT-5’.

Question: What would be the base sequence of RNA after transcription occurs? Turn this in.

(In this particular example, assume that the RNA product does not require processing to become mRNA. In other words, the transcribed RNA becomes the mRNA sequence.)

B) Step 2 – Translation: During protein synthesis at the ribosome, the base sequence of the mRNA codons is translated to the amino acid sequence of a protein.

Question: Using the mRNA that you transcribed above, use the genetic code table to determine the resulting amino acid sequence? Turn this in.

And, turn in the answer to these questions:

What is the significance of the first and last codons? What meaning do these codons have for protein synthesis?

C) Mutations: A mutation is defined as a change in the base sequence of DNA. This may occur as a “mistake” in DNA replication, for example.

Suppose that during DNA replication, two mutant DNA sequences are produced as shown below.

For the 2 mutated DNA sequences, you will investigate how these changes might affect the sequence of amino acids in a protein.

Question: For each of the two, you will need to first transcribe the mRNA, and then use the genetic code table to determine the amino acid sequence.

Turn these in, and state whether the protein sequence changes for each.

Question: Then, explain why a change in amino acid sequence might affect protein function. Turn in your answer.

Here is the original sequence, followed by two mutated sequences, 1 and 2:

Original sequence 3′- TACCCTTTAGTAGCCACT-5’

Mutated sequence 1) 3’-TACGCTTTAGTAGCCATT-5′

Mutated sequence 2) 3’-TAACCTTTACTAGGCACT-5’.

Part 2: Inheritance of Traits or Genetic Disorders (Chapter 12)

Bob and Sally recently married. Upon deciding to plan a family, both Sally and Bob find out that they are both heterozygous for cystic fibrosis, but neither of them has symptoms of the disorder.

Set up and complete a Punnett Square for cystic fibrosis for this couple; turn in the Punnett square.

When doing the Punnett Square, C = normal allele; and c = allele for cystic fibrosis.

Note: You can use the Table function in MS Word to create and fill in a Punnett Square.

Questions:

Based on the Punnett square, calculate chances (percentages) for having a healthy child (not a carrier), a child that is a carrier for the cystic fibrosis trait, and a child with cystic fibrosis? Turn in these percentages.

Part 3: Cell division, sexual reproduction and genetic variability (Chapter 11)

Eukaryotic cells can divide by mitosis or meiosis. In humans, mitosis produces new cells for growth and repair; meiosis produces sex cells (gametes) called sperm and eggs.

Although mutations are the ultimate source of genetic variability, both meiosis and sexual reproduction also can contribute to new genetic combinations in offspring.

Question: How do both meiosis and sexual reproduction (fertilization) produce offspring that differ genetically from the parents? Be sure to talk about steps in meiosis that increase variability as well as the process of fertilization.

Most detailed answer:

Answer by Kendi Rosenberg
part one:
A) AUGGGAAAUCAUCGGUGA
B) AUG(the start code/ methoninie), GGA (Glycine), AAU (Asparagine), CAU (Histidine), CGG (Arginine), UGA (Stop code)

The significance of the fist and last codons are that theay are the start code and the stop code.

The meaning they have for protien synthesis is that the start code tells where to start synthesis and the stop code tells where to stop synthesis.

C) Mutated sequence 1) AUGCGAAAUCAUCGGUAA
AUG (start code/ methonine), CGA (Arginine), AAU (Asparagine), CAU (Histidine), CGG (Arginine), UAA (Stop code)

Mutated sequence 2) AUU GGA AAU GAU CCG UGA
AUU (Lisoleucine), GGA (Glycine), AAU (Asparagine), GAU (Aspartic acid), CCG (Proline), UGA (Stop code)

The Change in Amino Acid sequence might effect the protien function because for example in mutated sequence number 2 there is no start code meaning this would never get translated.

Part 2:

Here i drew a punnet square for you and took a picture of it… here is the link

They have a 25% chance of having a child that is not a carrier.
They have a 50% chance having a child who is a carrier.
They have a 25% chance of having a child has cystic fibrosis
1:2:1 – genotypic
3:1- phenotypic

Part 3:

Meiosis and sexual reproduction contribute to genetic variation by independant assortment- random distribution of homologus churomosomes. Also by crossing over and random fertalization.

How about adding your own answer to the comments below!

Technorati Tags: Chapter, from, Genes, GENETICS, Mutations, part, Proteins