Plants with genotypes RRTT and rrtt were mated to produce an F1 plant. The F1 plant was testcrossed to a plant with genotype rrtt. Following are the numbers and types of progeny from this testcross: 38 RrTt 42 rrtt 8 Rrtt 12 rrTt
Question 1) What is the genetic distance between the R and T genes? Answer a. 10% b. 20% c. 40% d. 50% e. 80% Question 2) What is the genotype of the F1 individual? Answer a. RRTT b. RrTt c. Rrtt d. rrtt
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Answer to question no 2 will be RrTt
Not sure about question 1
ok thanks, i'll be sure to look over it and make sure it's right. I just wanted to make sure, to get the same answer so I'm correct.
can u help @sarah786
@Somy
oh im sorry i don't really know this @Abhisar
its okay...thnx for bothering @Somy
@Abhisar no problem)
genetic differences .lemme check O.o
ok \(\huge\ddot\smile\)
Really that term new to me . 2nd one is RrTt @shrutipande9 @Frostbite
@Koikkara
okay..it will be c
In genetic mapping different genes are linearly arranged according to their % of crossing over or recombination frequency or cross over frequency. Genes have different recombination frequencies depending on how closely linked they are. The genetic distance between genes is measured in map units (mu), where 1 mu is defined as the interval in which a 1% crossing over takes place. It is not a unit of physical distance like nanometers, but an indirect measure of distance on the basis of a probability. The higher this probability, the further the genes are from one another physically. The frequency of recombination is proportional to the distance between genes. The first genetic map ever constructed was based on recombination frequencies from Drosophila crosses involving the X-linked genes w (white eyes), m (miniature wings), and y (yellow body). In each cross, the number of parental progeny was much greater than the number of recombinant progeny. The percentages of recombinants were calculated for each cross. The recombination frequencies were: w-m – 32.7 % [which means that probability of coming together of w & m genes was 32.7 %] w-y – 1.0 % [which means that probability of coming together of w & y genes was 1.0 %] m-y – 33.7 % [which means that probability of coming together of m & y genes was 33.7 %] Chromosome: y__w___________________m - 1.0 mu - - 32.7 mu – Thus y and W are closest together since they have the lowest frequency of recombination. We also know that the m-w distance is shorter than the m-y distance, since m-y has a higher frequency of recombination than does w-m.
@aaronq can u pls chk this ?
oh wait the answer for q 1 is it 40%? @Abhisar
because if so it makes sense as the next generation are 38 RrTt 42 rrtt 8 Rrtt 12 rrTt the total number will be ur 100% when u add up 38+42+8+12= 100 number offspring in total that means RrTt will be 38% meaning approximately 40% rrtt will be 42% also around 40% Rrtt will be 8% approximately 10% rrTt will be 12% also around 10% you can round off , most important thing is that in total u have to get 100% after rounding so 40+40+10+10 will give you 100% thus % chances that R will come with T is 40% sorry @Abhisar i just didn't get the 'distance thing so i thought its something new
i still don't get 'distance' concept since i learnt this as % chances @Abhisar
this is according to ur info on distance concept, i hope u get it @Abhisar if not just let me know i'll explain
@Zale101
Q2 is right, good job boys @somy @Abhisar So for Q1 RT is dominant and therfore it is a apart of the original chromosomes, meaning that it will appear together most often = when looking at map units the closer to 1% mean that they are close to the same chromosomes. To figure this out we look at the parental which is either RT or rt, if it is a different combination we add up the columns. since 8 Rrtt 12 rrTt are not corresponding to rt, or RT we include them into our equation recombinant = # of recombinant / total 20/100 = .2 x 100% = 20% meaning that R T is 20% centimorgan apart from each other.
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