Biology Forum › Genetics › Can someone explain this to me???
March 16, 2010 at 7:44 am #12953sharingan23Participant
Ok, so I have the answers to these questions. Could someone please explain, why they are the answers? The answers are at the end of the question in the (brackets)
1) Black fur in mice (B) is dominant to brown fur (b). Short tails (T) are dominant to long tails (t). What fraction of the progeny of the cross BbTt × BBtt will have black fur and long tails?: (1/2)
2) Two true-breeding stocks of pea plants are crossed. One parent has red, axial flowers and the other has white, terminal flowers; all F1 individuals have red, axial flowers. The genes for flower color and location assort independently. If 1,000 F2 offspring resulted from the cross, approximately how many of them would you expect to have red, terminal flowers ? (190)
A man with type A blood marries a woman with type B blood. Their child has type O blood. What are the genotypes of the parents ? (Man IAi; woman IBi)
In cattle, roan coat color (mixed red and white hairs) occurs in the heterozygous (Rr) offspring of red (RR) and white (rr) homozygotes. Which of the following crosses would produce offspring in the ratio of 1 red : 2 roan : 1 white? (roan x roan)
In snapdragons, heterozygotes for one of the genes have pink flowers, whereas homozygotes have red or white flowers. When plants with red flowers are crossed with plants with white flowers, what proportion of the offspring will have pink flowers? (100%)
When crossing an organism that is homozygous recessive for a single trait with a heterozygote, what is the chance of producing an offspring with the homozygous recessive phenotype? (50%)
March 16, 2010 at 8:18 am #98311mamoruParticipant
Do you know how to do a Punnett Square to calculate the expected ratios of a genetic cross?
Because, if you do that for each example, you will get your answers. Do a dihybrid cross for numbers 1 and 2. Number 3 is probably answered in your textbook under the bloodtypes example of multiple alleles. If not, it will at least list all of the possible combinations to yield each bloodtype. If not, it should be relatively simple to do your own Punnett squares to figure them out. Number 4, do a monohybrid cross of each possible combination and the answer will be clear. Number 5, same thing. Number 6 as well.
If you don’t know how to do a Punnett square, read the wikipedia article I just linked and/or your chapter on Mendelian Genetics in your textbook. Try to work out the problems, and if you still have difficulty, post back what you have tried and specifically where you are confused.
March 16, 2010 at 9:09 am #98312sharingan23Participant
For number two. I don’t get how it’s 190. Should it not be 250???
For the snap dragons question, I’m getting 50 %
March 16, 2010 at 9:32 am #98313mamoruParticipantquote sharingan23:For number two. I don’t get how it’s 190. Should it not be 250???
How do you figure? Red and Axial are both dominant traits, and all members of the F1 generation are heterozygous. If you to a diyhybrid cross of heterozygous genotypes, you end up with a 9:3:3:1 ratio of the for possible phenotypes. Look at the Dihybrid cross section of the Punnett Square article I linked. Imagine that R = red, r = white, Y = axial, y = terminal. In the resulting cross, how many genotypes would show the Red/terminal phenotype? How many total squares are there in that cross? Divide the first number by the second number and multiply times 1000 (the number of plants in the F2).quote :For the snap dragons question, I’m getting 50 %
What is the genotype for Red flowers? What is the genotype for white flowers? Do a monhybrid cross of those genotypes and tell me what all possible resulting genotypes are. 😉
March 17, 2010 at 7:10 am #98329JackBeanParticipant
3) just think about it as: the child has 0, so it must be 00, the parents have A or B, respectively, so they must have at least one of these alleles, but they also must have 0 each to be able to pass it to the kid
March 18, 2010 at 3:58 pm #98369jwalinParticipant
everything sorted out over here…
3) simplified although jackbean has done well enough 😆
IaIo and IbIo are the genotypes
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