Dr. Bella towards the end of the video there's a question regarding U and T genetic condition, I think that it cant be dominant anyhow since the effective gene has skipped a generation ....see , none among P and q have condition....so it has to be recessive since we know that a dominant trait never skips a generation. On the other hand since no environmental conditions habe been mentioned therfore S can have mutation in egg....... Because only X given by father cant bring confition to U so one X from mother might have a recessive allele.....which futher adds up to that S is not diseased. According to me option C should be correct and not D....... Whatever may happen there's no sign of it being dominant. Am I wright?
Dr. Bella the cat question should have an answer as only 1, since 37°C is not less and in the question its given that greater the temperature lesser will be colour be, I mean its tone. ????
Even the question for heterozygotes in mice, it should be 8. Let me explain..... In first cross there is a black mouse and a white one and all children are black that means black parent was homozygous all all children were Bw if black parent is BB and white parent is ww. So here are 4 heterozygotes formed. 2 nd cross is done between two black mouse . Through previous cross we know that black mouse was BB so here also all children are black that means other black parent mouse should be Bb since, max heterozygous has to be found. Now here there are 2 BB and 2Bb that means now 2 heterozygotes are formed. Right? In the last cross between one black mouse and other white mouse we have one black parent envolved from 2 nd cross who is having allels Bb now here 2 black children and 2 white children are formed so white parent should be Ww, only then will Bw,BW,bw,bW be formed and here 2 heterozygotes are formed...... Adding up no. of heterozygotes from the beginning 4+2+2=8....
Please keep uploading such informative videos .God will surely bless u for this kind act.
Much informative ❤
Thank you for your help, please continue like that because it really helps.
Dr. Bella towards the end of the video there's a question regarding U and T genetic condition, I think that it cant be dominant anyhow since the effective gene has skipped a generation ....see , none among P and q have condition....so it has to be recessive since we know that a dominant trait never skips a generation.
On the other hand since no environmental conditions habe been mentioned therfore S can have mutation in egg....... Because only X given by father cant bring confition to U so one X from mother might have a recessive allele.....which futher adds up to that S is not diseased. According to me option C should be correct and not D.......
Whatever may happen there's no sign of it being dominant.
Am I wright?
Dr. Bella the cat question should have an answer as only 1, since 37°C is not less and in the question its given that greater the temperature lesser will be colour be, I mean its tone.
????
Concern this video hopefully you'll get your answer
ua-cam.com/video/EDo0m-AMv70/v-deo.htmlsi=vZKeLKZSwDjaL2xC
At 5:30 the correct option is C that's 8 heterozygous mice
Please look at your explanation and you will find your flaw
its actually 12 as it asked us the maximum number of mice that can be heterozygous black (theoretically)
Got it Miss Bella
Thanks for the explanation video ❤
Thank you! Does this have pdf
Even the question for heterozygotes in mice, it should be 8. Let me explain..... In first cross there is a black mouse and a white one and all children are black that means black parent was homozygous all all children were Bw if black parent is BB and white parent is ww. So here are 4 heterozygotes formed.
2 nd cross is done between two black mouse . Through previous cross we know that black mouse was BB so here also all children are black that means other black parent mouse should be Bb since, max heterozygous has to be found. Now here there are 2 BB and 2Bb that means now 2 heterozygotes are formed. Right?
In the last cross between one black mouse and other white mouse we have one black parent envolved from 2 nd cross who is having allels Bb now here 2 black children and 2 white children are formed so white parent should be Ww, only then will Bw,BW,bw,bW be formed and here 2 heterozygotes are formed......
Adding up no. of heterozygotes from the beginning 4+2+2=8....
Concern this link hopefully it'll clear your doubts
ua-cam.com/video/EDo0m-AMv70/v-deo.htmlsi=vZKeLKZSwDjaL2xC